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Proof step 1 from proof lim-exists (conclusion) (back to tree)

mon(σσ), σσ ≠ [] ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof lim-exists (previous step) (back to tree)

σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof lim-exists (previous step) (back to tree)

σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: # σσ = ∞

Additional information

Used formulas: # σσ = ∞

Proof step 4 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 5 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ∃ n. # σσ(n) = ∞

Additional information

Used formulas: ∃ n. # σσ(n) = ∞

Proof step 6 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ), ∃ n. # σσ(n) = ∞ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 7 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: σσ(n) abort: F discard: T computed: T

Additional information

Used terms: σσ(n)

Proof step 8 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(n) = ∞, σσ ≠ [], mon(σσ) ⊦ (∀ n0. σσ(n0) ⊑ σσ(n)) ∧ (∀ m. ¬ # σσ(n) < ⌜m⌝ → (∃ n0. σσ(n).take(m) ⊑ σσ(n0)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s)

Proof step 9 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(n) = ∞, σσ ≠ [], mon(σσ) ⊦ σσ(n0) ⊑ σσ(n) ∧ (∃ n0. σσ(n).take(m) ⊑ σσ(n0))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 10 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ ≠ [], mon(σσ), ¬ σσ(n0) ⊑ σσ(n) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:mon-lem, seq:mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n), subst:[σσ, m, n] → [σσ, n0, n]

Additional information

Used terms: σσ, n0, n

Proof step 11 from proof lim-exists (previous step) (back to tree)

mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

Lemma mon-lem

Proof step 12 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(n) = ∞, σσ ≠ [], mon(σσ), ¬ σσ(n0) ⊑ σσ(n) ⊦ n0 ≤ n ∧ ⌜n⌝ < # σσ

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ⌜m⌝ < ∞; (s)

Proof step 13 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ ≠ [], n < n0, mon(σσ), ¬ σσ(n0) ⊑ σσ(n) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:mon-lem, seq:mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n), subst:[σσ, m, n] → [σσ, n, n0]; [A] → [A]

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: σσ, n, n0

Proof step 14 from proof lim-exists (previous step) (back to tree)

mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

Lemma mon-lem

Proof step 15 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(n) = ∞, σσ ≠ [], n < n0, mon(σσ), σσ(n) ⊑ σσ(n0), ¬ σσ(n0) ⊑ σσ(n) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2); (s)

Proof step 16 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ(n) ≠ σσ(n0), σσ ≠ [], n < n0, mon(σσ), σσ(n) ⊑ σσ(n0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [σσ(n), σσ(n0)]; [A] → [A], <all paths>

Additional information

Used terms: σσ(n), σσ(n0)

Proof step 17 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 18 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(n) = ∞, σσ(n) ≠ σσ(n0), σσ ≠ [], n < n0, mon(σσ), ∃ σ1. σσ(n) + σ1 = σσ(n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 19 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ(n) + σ1 = σσ(n0), σσ(n) ≠ σσ(n0), σσ ≠ [], n < n0, mon(σσ) ⊦

User applied rule structural induction

And closed the goal!

Rule argument

term: σσ(n)

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜nf⌝ + σ = ⌜nf⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 20 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∃ n0. σσ(n).take(m) ⊑ σσ(n0)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n abort: F discard: T computed: F

Additional information

Used terms: n

Proof step 21 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(n) = ∞, σσ ≠ [], mon(σσ), ¬ σσ(n).take(m) ⊑ σσ(n) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [σσ(n).take(m), σσ(n)]; [A] → [A], <all paths>

Additional information

Used terms: σσ(n).take(m), σσ(n)

Proof step 22 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 23 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ∞, # σσ = ∞, σσ ≠ [], mon(σσ), ∀ σ1. σσ(n).take(m) + σ1 ≠ σσ(n) ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: σσ(n).drop(m) abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ σ.take(n) + σ.drop(n) = σ ↔ true; (ss) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old Used terms: σσ(n).drop(m)

Proof step 24 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ∀ n. ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

Additional information

Used formulas: ∀ n. ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

Proof step 25 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ), ∀ n. ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule choice left

And obtained the following new goals

Rule argument

pos: left 4

Additional information

Proof step 26 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: infinite-append-old, name:infinite-append, seq:⊦ ∃ σ, ii. ii(0) = 0 ∧ (∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ σ.drop(ii(n)).take(# lf(n)) = ⌜lf(n)⌝ ) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n)) ;∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1) ), subst:[lf] → [f_Nat->List<A>]; [A] → [A]

Additional information

Used terms: f_Nat->List<A>

Proof step 27 from proof lim-exists (previous step) (back to tree)

⊦ ∃ σ, ii. ii(0) = 0 ∧ (∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ σ.drop(ii(n)).take(# lf(n)) = ⌜lf(n)⌝ ) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1))

A spec-lemma

Proof step 28 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∃ σ, ii. ii(0) = 0 ∧ (∀ n. ii(n + 1) = ii(n) + # f_Nat->List<A>(n) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ σ.drop(ii(n)).take(# f_Nat->List<A>(n)) = ⌜f_Nat->List<A>(n)⌝ ) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 29 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), (finp(σ) ⊃ # σ = ⌜nnf(n)⌝ ∧ (∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n));∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1)), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ ¬ # σ < ⌜nnf(n + 1)⌝ ∧ σ.drop(nnf(n)).take(# f_Nat->List<A>(n)) = ⌜f_Nat->List<A>(n)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 5

Additional information

Proof step 30 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σ = ⌜nnf(n)⌝ , nnf(0) = 0, σσ ≠ [], finp(σ), mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ ¬ # σ < ⌜nnf(n + 1)⌝ ∧ σ.drop(nnf(n)).take(# f_Nat->List<A>(n)) = ⌜f_Nat->List<A>(n)⌝ , ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s) ⊦ ⌜x⌝ .take(n) = ⌜x.take(n)⌝ ; (s) ⊦ ⌜x⌝ .drop(n) = ⌜x.drop(n)⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s) ⊦ finp(σ) ↔ (∃ x. ⌜x⌝ = σ); (as)

Proof step 31 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n abort: F discard: F computed: F

Additional information

Used terms: n

Proof step 32 from proof lim-exists (previous step) (back to tree)

# x = nnf(n), # σσ = ∞, nnf(0) = 0, # σσ(n) ≠ ∞, σσ ≠ [], mon(σσ), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-inf, name: notinf-fin, seq: ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ ), subst:[n∞] → [# σσ(n)], <all paths>

Additional information

Used terms: # σσ(n)

Proof step 33 from proof lim-exists (previous step) (back to tree)

⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

A spec-lemma

Proof step 34 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∃ n0. # σσ(n) = ⌜n0⌝ ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 35 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ∀ m. n ≤ m → σσ(m) = σσ(n)

Additional information

Used formulas: ∀ m. n ≤ m → σσ(m) = σσ(n)

Proof step 36 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → σσ(m) = σσ(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 2 terms: σσ(n) abort: F discard: T computed: T

Additional information

Used terms: σσ(n)

Proof step 37 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ m. n ≤ m → σσ(m) = σσ(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞, (∀ n0. σσ(n0) ⊑ σσ(n)) ∧ (∀ m. ¬ # σσ(n) < ⌜m⌝ → (∃ n0. σσ(n).take(m) ⊑ σσ(n0)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

Proof step 38 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ m. n ≤ m → σσ(m) = σσ(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞, σσ(n1) ⊑ σσ(n) ∧ (¬ n0 < m → (∃ n0. σσ(n).take(m) ⊑ σσ(n0)))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Proof step 39 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ σσ(n1) ⊑ σσ(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → σσ(m) = σσ(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 12 terms: n1 abort: F discard: F computed: F

Additional information

Used terms: n1

Proof step 40 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ σσ(n1) ⊑ σσ(n), n ≤ n1 → σσ(n1) = σσ(n), ∀ m. n ≤ m → σσ(m) = σσ(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:mon-lem, seq:mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n), subst:[σσ, m, n] → [σσ, n1, n]

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → m ≤ n; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) ⊦ σ ⊑ σ; (s) Used terms: σσ, n1, n

Proof step 41 from proof lim-exists (previous step) (back to tree)

mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

Lemma mon-lem

Proof step 42 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ n0 < m, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → σσ(m) = σσ(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ n0. σσ(n).take(m) ⊑ σσ(n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 43 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), m ≤ n0, ∀ m. n ≤ m → σσ(m) = σσ(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n0. σσ(n).take(m) ⊑ σσ(n0), ∃ n. # σσ(n) = ∞

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n abort: F discard: T computed: F

Additional information

Used terms: n

Proof step 44 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), m ≤ n0, ¬ σσ(n).take(m) ⊑ σσ(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → σσ(m) = σσ(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [σσ(n).take(m), σσ(n)]; [A] → [A], <all paths>

Additional information

Used terms: σσ(n).take(m), σσ(n)

Proof step 45 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 46 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), m ≤ n0, ∀ m. n ≤ m → σσ(m) = σσ(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ σ1. σσ(n).take(m) + σ1 ≠ σσ(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And closed the goal!

Rule argument

pos: left 13 terms: σσ(n).drop(m) abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) ¬ # σ < ⌜n⌝ ⊦ σ.take(n) + σ.drop(n) = σ ↔ true; (ss) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old Used terms: σσ(n).drop(m)

Proof step 47 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∀ m. n ≤ m → σσ(m) = σσ(n), ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 48 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ(m) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m, ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule weakening

And obtained the following new goals

Rule argument

poslist: right 1

Additional information

Proof step 49 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ(m) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∃ n. # σσ(n) = ∞

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: m

Additional information

Used simplifier rules: Nat freely generated by 0, +1; ⊦ m ≤ 0 ↔ m = 0; (s)

Proof step 50 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ(m0.+1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0.+1, ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ σσ, x, nnf, n, n0, f_Nat->List<A>. # σσ = ∞ ∧ # x = nnf(n) ∧ # σσ(n) = ⌜n0⌝ ∧ nnf(0) = 0 ∧ σσ(m0) ≠ σσ(n) ∧ σσ ≠ [] ∧ mon(σσ) ∧ n ≤ m0 ∧ (∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ) ∧ (∀ m, n. m ≤ n → nnf(m) ≤ nnf(n)) ∧ (∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n)) ∧ (∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0)) → (∃ n. # σσ(n) = ∞) ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s)

Proof step 51 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0 + 1, ∀ σσ, x, nnf, n, n0, f_Nat->List<A>. # σσ = ∞ ∧ # x = nnf(n) ∧ # σσ(n) = ⌜n0⌝ ∧ nnf(0) = 0 ∧ σσ(m0) ≠ σσ(n) ∧ σσ ≠ [] ∧ mon(σσ) ∧ n ≤ m0 ∧ (∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ) ∧ (∀ m, n. m ≤ n → nnf(m) ≤ nnf(n)) ∧ (∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n)) ∧ (∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0)) → (∃ n. # σσ(n) = ∞), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 9 terms: σσ, x, nnf, n, n0, f_Nat->List<A> abort: F discard: T computed: T

Additional information

Used terms: σσ, x, nnf, n, n0, f_Nat->List<A>

Proof step 52 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, nnf(0) = 0, σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0 + 1, # σσ = ∞ ∧ # x = nnf(n) ∧ # σσ(n) = ⌜n0⌝ ∧ nnf(0) = 0 ∧ σσ(m0) ≠ σσ(n) ∧ σσ ≠ [] ∧ mon(σσ) ∧ n ≤ m0 ∧ (∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ) ∧ (∀ m, n. m ≤ n → nnf(m) ≤ nnf(n)) ∧ (∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n)) ∧ (∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0)) → (∃ n. # σσ(n) = ∞), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 53 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , nnf(0) = 0, σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0 + 1, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∃ n. # σσ(n) = ∞, σσ(m0) ≠ σσ(n) ∧ n ≤ m0

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 12 terms: m0 abort: F discard: F computed: F

Additional information

Used terms: m0

Proof step 54 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, x.drop(nnf(m0)).take(# f_Nat->List<A>(m0)) = f_Nat->List<A>(m0), nnf(0) = 0, nnf(m0 + 1) = nnf(m0) + # f_Nat->List<A>(m0), σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0 + 1, ¬ nnf(n) < nnf(m0 + 1), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞, σσ(m0) ≠ σσ(n) ∧ n ≤ m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 55 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , x.drop(nnf(m0)).take(# f_Nat->List<A>(m0)) = f_Nat->List<A>(m0), nnf(m0 + 1) = nnf(m0) + # f_Nat->List<A>(m0), nnf(0) = 0, σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0 + 1, nnf(m0 + 1) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∃ n. # σσ(n) = ∞, σσ(m0) ≠ σσ(n) ∧ n ≤ m0

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 12 terms: m0 abort: F discard: F computed: T

Additional information

Used terms: m0

Proof step 56 from proof lim-exists (previous step) (back to tree)

# σσ(n) = ⌜n0⌝ , # x = nnf(n), # σσ = ∞, x.drop(nnf(m0)).take(# f_Nat->List<A>(m0)) = f_Nat->List<A>(m0), σσ(m0 + 1) = σσ(m0) + ⌜f_Nat->List<A>(m0)⌝ , nnf(0) = 0, nnf(m0 + 1) = nnf(m0) + # f_Nat->List<A>(m0), σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), nnf(m0 + 1) ≤ nnf(n), n ≤ m0 + 1, ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞, σσ(m0) ≠ σσ(n) ∧ n ≤ m0

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 14 terms: m0 abort: F discard: F computed: F

Additional information

Used terms: m0

Proof step 57 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # x = nnf(n), # σσ(n) = ⌜n0⌝ , x.drop(nnf(m0)).take(# f_Nat->List<A>(m0)) = f_Nat->List<A>(m0), σσ(m0 + 1) = σσ(m0) + ⌜f_Nat->List<A>(m0)⌝ , nnf(m0 + 1) = nnf(m0) + # f_Nat->List<A>(m0), nnf(0) = 0, σσ(m0 + 1) ≠ σσ(n), σσ ≠ [], mon(σσ), n ≤ m0 + 1, nnf(m0 + 1) ≤ nnf(n), n ≤ m0 → f_Nat->List<A>(m0) = [] ∧ nnf(m0) = nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ m. n ≤ m → f_Nat->List<A>(m) = [] ∧ nnf(m) = nnf(n), ∀ n0. nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0) ∧ ¬ nnf(n) < nnf(n0 + 1) ∧ x.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0) ⊦ ∃ n. # σσ(n) = ∞, σσ(m0) ≠ σσ(n) ∧ n ≤ m0

User applied rule cut formula

And closed the goal!

Rule argument

fma: n ≤ m0

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ # x = n → x.drop(n) = []; (s) ⊦ x.take(0) = []; (s) ⊦ # [] = 0; (s) ⊦ n + 0 = n; (s) ⊦ ⌜[]⌝ = []; (s) ⊦ σ + [] = σ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old Used formulas: n ≤ m0

Proof step 58 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ finp(σ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ ¬ # σ < ⌜nnf(n + 1)⌝ ∧ σ.drop(nnf(n)).take(# f_Nat->List<A>(n)) = ⌜f_Nat->List<A>(n)⌝ , ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s) ⊦ ⌜nf⌝ .take(n) = ⌜nf.take(n)⌝ ; (s) ⊦ ⌜nf⌝ .drop(n) = ⌜nf.drop(n)⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ finp(σ) ↔ ¬(∃ nf. ⌜nf⌝ = σ); (ss)

Proof step 59 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝

Additional information

Used formulas: ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝

Proof step 60 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: σσ(0) + ⌜nf⌝ abort: F discard: T computed: F

Additional information

Used terms: σσ(0) + ⌜nf⌝

Proof step 61 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, (∀ n. σσ(n) ⊑ σσ(0) + ⌜nf⌝ ) ∧ (∀ m. ¬ #(σσ(0) + ⌜nf⌝ ) < ⌜m⌝ → (∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ n∞ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 62 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. # σσ(n) = ∞, σσ(n) ⊑ σσ(0) + ⌜nf⌝ ∧ (∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Proof step 63 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ σσ(n) ⊑ σσ(0) + ⌜nf⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 6 terms: n abort: F discard: F computed: F

Additional information

Used terms: n

Proof step 64 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ σσ(n) ⊑ σσ(0) + ⌜nf⌝ , ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 2, flags: discard

Additional information

Proof step 65 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ σσ(0) + ⌜nf.take(nnf(n))⌝ ⊑ σσ(0) + ⌜nf⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [σσ(0) + ⌜nf.take(nnf(n))⌝ , σσ(0) + ⌜nf⌝ ]; [A] → [A], <all paths>

Additional information

Used terms: σσ(0) + ⌜nf.take(nnf(n))⌝ , σσ(0) + ⌜nf⌝

Proof step 66 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 67 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ σ1. (σσ(0) + ⌜nf.take(nnf(n))⌝ ) + σ1 ≠ σσ(0) + ⌜nf⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: ⌜nf.drop(nnf(n))⌝ abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ σ + σ1 = σ + σ2 ↔ σ1 = σ2 ∨ # σ = ∞; (ws) ⊦ ⌜nf.take(n)⌝ + ⌜nf.drop(n)⌝ = ⌜nf⌝ ↔ true; (ss) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old Used terms: ⌜nf.drop(nnf(n))⌝

Proof step 68 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: 0 abort: F discard: F computed: F

Additional information

Used terms: 0

Proof step 69 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, # σσ(0) ≠ ∞, σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n), ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-inf, name: notinf-fin, seq: ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ ), subst:[n∞] → [# σσ(0)], <all paths>

Additional information

Used terms: # σσ(0)

Proof step 70 from proof lim-exists (previous step) (back to tree)

⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

A spec-lemma

Proof step 71 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∃ n. # σσ(0) = ⌜n⌝ ⊦ ∃ n. # σσ(n) = ∞, ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 72 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n), ∃ n. # σσ(n) = ∞

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: m < n

Additional information

Used formulas: m < n

Proof step 73 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , nnf(0) = 0, σσ ≠ [], m < n, mon(σσ), ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 2 terms: 0 abort: F discard: T computed: F

Additional information

Used terms: 0

Proof step 74 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, nnf(0) = 0, σσ ≠ [], m < n, mon(σσ), ¬ (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(0), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [(σσ(0) + ⌜nf⌝ ).take(m), σσ(0)]; [A] → [A], <all paths>

Additional information

Used terms: (σσ(0) + ⌜nf⌝ ).take(m), σσ(0)

Proof step 75 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 76 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , nnf(0) = 0, σσ ≠ [], m < n, mon(σσ), ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ σ1. (σσ(0) + ⌜nf⌝ ).take(m) + σ1 ≠ σσ(0) ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 12 terms: σσ(0).drop(m) abort: F discard: T computed: F

Additional information

Used terms: σσ(0).drop(m)

Proof step 77 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, nnf(0) = 0, (σσ(0) + ⌜nf⌝ ).take(m) + σσ(0).drop(m) ≠ σσ(0), σσ ≠ [], m < n, mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: append-to-ls, seq: ⊦ ¬ # σ < ⌜n⌝ → (σ + σ0).take(n) = σ.take(n), subst:[σ, σ0, n] → [σσ(0), ⌜nf⌝ , m]; [A] → [A], <all paths>

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m < n → ¬ n < m; (s) ¬ # σ < ⌜n⌝ ⊦ σ.take(n) + σ.drop(n) = σ ↔ true; (ss) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) Used terms: σσ(0), ⌜nf⌝ , m

Proof step 78 from proof lim-exists (previous step) (back to tree)

⊦ ¬ # σ < ⌜n⌝ → (σ + σ0).take(n) = σ.take(n)

A spec-lemma

Proof step 79 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), ¬ m < n, ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 80 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), n ≤ m, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n), ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 8 terms: m - n abort: F discard: T computed: F

Additional information

Used terms: m - n

Proof step 81 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , nnf(0) = 0, σσ ≠ [], mon(σσ), n ≤ m, ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∃ m0. nnf(m0) ≤ m - n ∧ m - n < nnf(m0 + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < m + n ↔ n ≠ 0; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s) ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)

Proof step 82 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, nnf(0) = 0, σσ ≠ [], nnf(m0) < nnf(m0 + 1), m < nnf(m0 + 1) + n, mon(σσ), n ≤ m, nnf(m0) + n ≤ m, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ ⊦ ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n), ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 13 terms: m0 abort: F discard: F computed: F

Additional information

Used terms: m0

Proof step 83 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, σσ ≠ [], m < nnf(m0 + 1) + n, nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) + n ≤ m, n ≤ m, ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n)

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 11 terms: m0 + 1 abort: F discard: T computed: F

Additional information

Used terms: m0 + 1

Proof step 84 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , σσ(m0 + 1) = σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , nnf(0) = 0, σσ ≠ [], nnf(m0) < nnf(m0 + 1), m < nnf(m0 + 1) + n, mon(σσ), n ≤ m, nnf(m0) + n ≤ m, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(n), ∃ n. # σσ(n) = ∞

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: m0 + 1 abort: F discard: T computed: F

Additional information

Used terms: m0 + 1

Proof step 85 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0 + 1) = σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, σσ ≠ [], m < nnf(m0 + 1) + n, nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) + n ≤ m, n ≤ m, ¬ (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(m0 + 1), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 3, flags: discard

Additional information

Proof step 86 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, σσ ≠ [], nnf(m0) < nnf(m0 + 1), m < nnf(m0 + 1) + n, mon(σσ), n ≤ m, nnf(m0) + n ≤ m, ¬ (σσ(0) + ⌜nf⌝ ).take(m) ⊑ σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [(σσ(0) + ⌜nf⌝ ).take(m), σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ ]; [A] → [A], <all paths>

Additional information

Used terms: (σσ(0) + ⌜nf⌝ ).take(m), σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝

Proof step 87 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 88 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, σσ ≠ [], m < nnf(m0 + 1) + n, nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) + n ≤ m, n ≤ m, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ σ1. (σσ(0) + ⌜nf⌝ ).take(m) + σ1 ≠ σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 14 terms: ⌜nf⌝ .take(nnf(m0 + 1)).drop(m - n) abort: F discard: T computed: F

Additional information

Used terms: ⌜nf⌝ .take(nnf(m0 + 1)).drop(m - n)

Proof step 89 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, (σσ(0) + ⌜nf⌝ ).take(m) + ⌜nf⌝ .take(nnf(m0 + 1)).drop(m - n) ≠ σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , σσ ≠ [], nnf(m0) < nnf(m0 + 1), m < nnf(m0 + 1) + n, mon(σσ), n ≤ m, nnf(m0) + n ≤ m, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .drop(n) = ⌜x.drop(n)⌝ ; (s) ⊦ ⌜nf⌝ .take(n) = ⌜nf.take(n)⌝ ; (s)

Proof step 90 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, (σσ(0) + ⌜nf⌝ ).take(m) + ⌜nf.take(nnf(m0 + 1)).drop(m - n)⌝ ≠ σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , σσ ≠ [], m < nnf(m0 + 1) + n, nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) + n ≤ m, n ≤ m, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [m, n, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, n, n1

Proof step 91 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , m = n1 + n, nnf(0) = 0, (σσ(0) + ⌜nf⌝ ).take(m) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ ≠ σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , σσ ≠ [], nnf(m0) < nnf(m0 + 1), m < nnf(m0 + 1) + n, mon(σσ), n ≤ m, nnf(m0) + n ≤ m, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m ≤ m + n; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)

Proof step 92 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, (σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ ≠ σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ , σσ ≠ [], n1 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: extensionality, seq: ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0(n) = σ1(n)), subst:[σ0, σ1] → [(σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ , σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ ]; [A] → [A], <all paths>

Additional information

Used terms: (σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ , σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝

Proof step 93 from proof lim-exists (previous step) (back to tree)

⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0(n) = σ1(n))

A spec-lemma

Proof step 94 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, σσ ≠ [], nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , #((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ ) ≠ #(σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ ) ∨ (∃ n0. ¬ ( ⌜n0⌝ < #((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ ) → ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) = (σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ )(n0))) ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n < m → (m - n) + n = m; (s) ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s) ⊦ ⌜n1⌝ < ⌜n0⌝ + ⌜n⌝ + ⌜m⌝ ↔ n1 < n0 + n + m; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ # nf.take(n) = n; (s) ¬ # x < n ⊦ # x.drop(n) = # x - n; (s) ⊦ m < n → ¬ n < m; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) ⊦ n∞ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # ⌜nf.take(n)⌝ = ⌜n⌝ ; (s)

Proof step 95 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ (σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ )(n0), σσ ≠ [], nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), n0 < nnf(m0 + 1) + n, mon(σσ), nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: get-append-ls, seq: ⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m), subst:[σ, σ1, m] → [σσ(0), ⌜nf.take(nnf(m0 + 1))⌝ , n0]; [A] → [A], <all paths>

Additional information

Used terms: σσ(0), ⌜nf.take(nnf(m0 + 1))⌝ , n0

Proof step 96 from proof lim-exists (previous step) (back to tree)

⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m)

A spec-lemma

Proof step 97 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ (σσ(0) + ⌜nf.take(nnf(m0 + 1))⌝ )(n0), σσ ≠ [], n0 < nnf(m0 + 1) + n, n1 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, ¬ ⌜n0⌝ < # σσ(0), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ m < n → (n ≤ m ↔ false); (s) n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ⌜n⌝ .n = n; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ⌜x⌝ (n) = x(n); (s) ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)(m) = σ1(m - (# σ).n); (s)

Proof step 98 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ nf(n0 - n), σσ ≠ [], n0 < nnf(m0 + 1) + n, n1 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, n ≤ n0, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [n0, n, n2]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n, n2

Proof step 99 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , n0 = n2 + n, nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ nf(n2), σσ ≠ [], nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), n0 < nnf(m0 + 1) + n, mon(σσ), n ≤ n0, nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m ≤ m + n; (s)

Proof step 100 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n2 + n) ≠ nf(n2), σσ ≠ [], n1 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), n2 < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: get-append-ls, seq: ⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m), subst:[σ, σ1, m] → [(σσ(0) + ⌜nf⌝ ).take(n1 + n), ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ , n2 + n]; [A] → [A], <all paths>

Additional information

Used terms: (σσ(0) + ⌜nf⌝ ).take(n1 + n), ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ , n2 + n

Proof step 101 from proof lim-exists (previous step) (back to tree)

⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m)

A spec-lemma

Proof step 102 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n2 + n) ≠ nf(n2), σσ ≠ [], n2 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, ¬ ⌜n2 + n⌝ < # (σσ(0) + ⌜nf⌝ ).take(n1 + n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ m < n → (n ≤ m ↔ false); (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n < m → (m - n) + n = m; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m ≤ n → (n - m = 0 ↔ n = m); (s) ⊦ m < m + n ↔ n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) m + m0 < # x ⊦ x.drop(m)(m0) = x(m + m0); (s) ⊦ # nf.take(n) = n; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ + ⌜n1⌝ ↔ n + m < n0 + n1; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ n∞ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ ⌜x⌝ (n) = x(n); (s) ⊦ m + n - n + n1 = m - n1; (s) ⊦ ⌜n⌝ .n = n; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)(m) = σ1(m - (# σ).n); (s)

Proof step 103 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, nf(n1 + (n2 - n1)) ≠ nf(n2), σσ ≠ [], nnf(m0) < nnf(m0 + 1), n2 < nnf(m0 + 1), n1 < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, n1 ≤ n2, nnf(m0) ≤ n2, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [n2, n1, n3]

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, n1, n3

Proof step 104 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, (σσ(0) + ⌜nf⌝ ).take(n1 + n)(n2 + n) ≠ nf(n2), σσ ≠ [], n2 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), ⌜n2 + n⌝ < # (σσ(0) + ⌜nf⌝ ).take(n1 + n), mon(σσ), nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) trans-Nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic m < n, ¬ # σ < ⌜n⌝ ⊦ σ.take(n)(m) = σ(m); (s) ⊦ ¬ ∞ < n∞; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ n∞ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ + ⌜n1⌝ ↔ n + m < n0 + n1; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s)

Proof step 105 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, (σσ(0) + ⌜nf⌝ )(n2 + n) ≠ nf(n2), σσ ≠ [], n2 < n1, n2 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), n2 + 1 < nnf(m0 + 1), mon(σσ), nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: get-append-ls, seq: ⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m), subst:[σ, σ1, m] → [σσ(0), ⌜nf⌝ , n2 + n]; [A] → [A], <all paths>

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)(m) = σ1(m - (# σ).n); (s) ⊦ ⌜nf⌝ (n) = nf(n); (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ⌜n⌝ .n = n; (s) ⊦ m + n - m = n; (s) Used terms: σσ(0), ⌜nf⌝ , n2 + n

Proof step 106 from proof lim-exists (previous step) (back to tree)

⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m)

A spec-lemma

Proof step 107 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ σσ(0)(n0), σσ ≠ [], n0 < nnf(m0 + 1) + n, n1 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), ⌜n0⌝ < # σσ(0), mon(σσ), nnf(m0) ≤ n1, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

Proof step 108 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ σσ(0)(n0), σσ ≠ [], n0 < n, nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), n0 < nnf(m0 + 1) + n, mon(σσ), nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: get-append-notls, seq: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)(m) = σ1(m - (# σ).n), subst:[σ, σ1, m] → [(σσ(0) + ⌜nf⌝ ).take(n1 + n), ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ , n0]; [A] → [A], <all paths>

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s) ⊦ ⌜n⌝ .n = n; (s) ⊦ ⌜x⌝ (n) = x(n); (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ n∞ + ∞ = ∞; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → ⌜m⌝ < ⌜n⌝ + n∞0; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf Used terms: (σσ(0) + ⌜nf⌝ ).take(n1 + n), ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ , n0

Proof step 109 from proof lim-exists (previous step) (back to tree)

⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)(m) = σ1(m - (# σ).n)

A spec-lemma

Proof step 110 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, # σσ(0) = ⌜n⌝ , σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, ((σσ(0) + ⌜nf⌝ ).take(n1 + n) + ⌜nf.take(nnf(m0 + 1)).drop(n1)⌝ )(n0) ≠ σσ(0)(n0), σσ ≠ [], ⌜n0⌝ < # (σσ(0) + ⌜nf⌝ ).take(n1 + n), n0 < nnf(m0 + 1) + n, n1 < nnf(m0 + 1), nnf(m0) < nnf(m0 + 1), n0 < n, mon(σσ), nnf(m0) ≤ n1, ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ m < n → ⌜m⌝ < ⌜n⌝ + n∞0; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ ¬ ∞ < n∞; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ n∞ + ∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) m < n, ¬ # σ < ⌜n⌝ ⊦ σ.take(n)(m) = σ(m); (s) ⊦ ¬ m + n < m; (s) ⊦ m + n < m0 → m + n < m0 + n0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m); (s)

Proof step 111 from proof lim-exists (previous step) (back to tree)

# σσ(0) = ⌜n⌝ , # σσ = ∞, σσ(m0) = σσ(0) + ⌜nf.take(nnf(m0))⌝ , nnf(0) = 0, (σσ(0) + ⌜nf⌝ )(n0) ≠ σσ(0)(n0), σσ ≠ [], n0 < n, nnf(m0) < nnf(m0 + 1), n1 < nnf(m0 + 1), n0 < nnf(m0 + 1) + n, mon(σσ), nnf(m0) ≤ n1, ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: get-append-ls, seq: ⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m), subst:[σ, σ1, m] → [σσ(0), ⌜nf⌝ , n0]; [A] → [A], <all paths>

Additional information

Used simplifier rules: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)(m) = σ1(m - (# σ).n); (s) ⊦ ⌜nf⌝ (n) = nf(n); (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ⌜n⌝ .n = n; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: σσ(0), ⌜nf⌝ , n0

Proof step 112 from proof lim-exists (previous step) (back to tree)

⊦ ⌜m⌝ < # σ → (σ + σ1)(m) = σ(m)

A spec-lemma

Proof step 113 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∀ n. σσ(n) = σσ(0) + ⌜nf.take(nnf(n))⌝ , ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 114 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ(n) ≠ σσ(0) + ⌜nf.take(nnf(n))⌝ , σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: n

Additional information

Used simplifier rules: Nat freely generated by 0, +1; ⊦ σ + σ0 = σ ↔ σ0 = [] ∨ # σ = ∞; (s) ⊦ ⌜x⌝ = [] ↔ x = []; (s) ⊦ nf.take(n) = [] ↔ n = 0; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old

Proof step 115 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ(n0.+1) ≠ σσ(0) + ⌜nf.take(nnf(n0.+1))⌝ , σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ σσ, nnf, nf, f_Nat->List<A>. # σσ = ∞ ∧ nnf(0) = 0 ∧ σσ(n0) ≠ σσ(0) + ⌜nf.take(nnf(n0))⌝ ∧ σσ ≠ [] ∧ mon(σσ) ∧ (∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ) ∧ (∀ m, n. m ≤ n → nnf(m) ≤ nnf(n)) ∧ (∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1)) ∧ (∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n)) → (∃ n. # σσ(n) = ∞) ∨ (∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s)

Proof step 116 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ(n0 + 1) ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ σσ, nnf, nf, f_Nat->List<A>. # σσ = ∞ ∧ nnf(0) = 0 ∧ σσ(n0) ≠ σσ(0) + ⌜nf.take(nnf(n0))⌝ ∧ σσ ≠ [] ∧ mon(σσ) ∧ (∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ) ∧ (∀ m, n. m ≤ n → nnf(m) ≤ nnf(n)) ∧ (∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1)) ∧ (∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n)) → (∃ n. # σσ(n) = ∞) ∨ (∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 6 terms: σσ, nnf, nf, f_Nat->List<A> abort: F discard: T computed: T

Additional information

Used terms: σσ, nnf, nf, f_Nat->List<A>

Proof step 117 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, σσ(n0 + 1) ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), # σσ = ∞ ∧ nnf(0) = 0 ∧ σσ(n0) ≠ σσ(0) + ⌜nf.take(nnf(n0))⌝ ∧ σσ ≠ [] ∧ mon(σσ) ∧ (∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ) ∧ (∀ m, n. m ≤ n → nnf(m) ≤ nnf(n)) ∧ (∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1)) ∧ (∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n)) → (∃ n. # σσ(n) = ∞) ∨ (∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 118 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n0) = σσ(0) + ⌜nf.take(nnf(n0))⌝ , nnf(0) = 0, σσ(n0 + 1) ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ n0. σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 7 terms: n0 abort: F discard: T computed: T

Additional information

Used terms: n0

Proof step 119 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n0) = σσ(0) + ⌜nf.take(nnf(n0))⌝ , σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , nnf(0) = 0, σσ(n0 + 1) ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ n. nnf(n + 1) = nnf(n) + # f_Nat->List<A>(n) ∧ nf.drop(nnf(n)).take(# f_Nat->List<A>(n)) = f_Nat->List<A>(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 8 terms: n0 abort: F discard: T computed: F

Additional information

Used terms: n0

Proof step 120 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , σσ(n0) = σσ(0) + ⌜nf.take(nnf(n0))⌝ , nnf(0) = 0, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), σσ(n0 + 1) ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 121 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), σσ(n0) = σσ(0) + ⌜nf.take(nnf(n0))⌝ , σσ(n0 + 1) = σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ , nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), nnf(0) = 0, σσ(n0 + 1) ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 4, flags: discard

Additional information

Proof step 122 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), σσ(n0) = σσ(0) + ⌜nf.take(nnf(n0))⌝ , nnf(0) = 0, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), σσ(n0) + ⌜f_Nat->List<A>(n0)⌝ ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 3, flags: discard

Additional information

Proof step 123 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), nnf(0) = 0, (σσ(0) + ⌜nf.take(nnf(n0))⌝ ) + ⌜f_Nat->List<A>(n0)⌝ ≠ σσ(0) + ⌜nf.take(nnf(n0 + 1))⌝ , σσ ≠ [], mon(σσ), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z; (s) ⊦ σ + σ1 = σ + σ2 ↔ σ1 = σ2 ∨ # σ = ∞; (ws)

Proof step 124 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) = f_Nat->List<A>(n0), nnf(0) = 0, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), # σσ(0) ≠ ∞, nf.take(nnf(n0)) + f_Nat->List<A>(n0) ≠ nf.take(nnf(n0 + 1)), σσ ≠ [], mon(σσ), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 2, flags: rotate, discard

Additional information

Proof step 125 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(n0 + 1) = nnf(n0) + # nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)), nnf(0) = 0, # σσ(0) ≠ ∞, nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) ≠ nf.take(nnf(n0 + 1)), σσ ≠ [], mon(σσ), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ # nf.take(n) = n; (s)

Proof step 126 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), # σσ(0) ≠ ∞, nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)) ≠ nf.take(nnf(n0 + 1)), σσ ≠ [], mon(σσ), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-getput, name: extension, seq: ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x(n) = y(n)), subst:[y, x] → [nf.take(nnf(n0 + 1)), nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0))], <all paths>

Additional information

Used terms: nf.take(nnf(n0 + 1)), nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0))

Proof step 127 from proof lim-exists (previous step) (back to tree)

⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x(n) = y(n))

A spec-lemma

Proof step 128 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), nnf(0) = 0, # σσ(0) ≠ ∞, σσ ≠ [], mon(σσ), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), #(nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0))) ≠ # nf.take(nnf(n0 + 1)) ∨ (∃ n. ¬ ( n < #(nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0))) → (nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)))(n) = nf.take(nnf(n0 + 1))(n))) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ m ≤ n → (n < m ↔ false); (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ # nf.take(n) = n; (s) ⊦ #(x + y) = # x + # y; (s)

Proof step 129 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), nnf(0) = 0, # σσ(0) ≠ ∞, (nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)))(n) ≠ nf(n), σσ ≠ [], n < nnf(n0 + 1), mon(σσ), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del, name: get-append-less, seq: ⊦ n < # x → (x + y)(n) = x(n), subst:[n, y, x] → [n, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)), nf.take(nnf(n0))], <all paths>

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ # nf.take(n) = n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) Used terms: n, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)), nf.take(nnf(n0))

Proof step 130 from proof lim-exists (previous step) (back to tree)

⊦ n < # x → (x + y)(n) = x(n)

A spec-lemma

Proof step 131 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(0) = 0, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), # σσ(0) ≠ ∞, (nf.take(nnf(n0)) + nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0)))(n) ≠ nf(n), σσ ≠ [], n < nnf(n0 + 1), mon(σσ), ¬ n < # nf.take(nnf(n0)), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n), ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1) ⊦ ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ # x = 0 ↔ x = []; (s) ⊦ m < m + n ↔ n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ # nf.take(n) = n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ ¬ n < # x → (x + y)(n) = y(n - # x); (s)

Proof step 132 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, nnf(n0 + 1) = nnf(n0) + # f_Nat->List<A>(n0), nnf(0) = 0, # σσ(0) ≠ ∞, nf.drop(nnf(n0)).take(# f_Nat->List<A>(n0))(n - nnf(n0)) ≠ nf(n), f_Nat->List<A>(n0) ≠ [], σσ ≠ [], n < nnf(n0 + 1), nnf(n0) < nnf(n0 + 1), mon(σσ), nnf(n0) ≤ n, ∀ n. ∃ m. nnf(m) ≤ n ∧ n < nnf(m + 1), ∀ m, n. m ≤ n → nnf(m) ≤ nnf(n) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [n, nnf(n0), n1]

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≤ m + n; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m ≤ n → (n < m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ nf.drop(n)(m) = nf(m + n); (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, nnf(n0), n1

Proof step 133 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∀ n. ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ , ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 134 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞, ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: mon-def, seq: ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1)), subst:[σσ] → [σσ]; [A] → [A], <all paths>

Additional information

Used terms: σσ

Proof step 135 from proof lim-exists (previous step) (back to tree)

⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1))

Lemma mon-def

Proof step 136 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1) ⊦ ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ , ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)

Proof step 137 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], ∀ n0. σσ(n0) ⊑ σσ(n0 + 1) ⊦ ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ , ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: n abort: F discard: T computed: F

Additional information

Used terms: n

Proof step 138 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], σσ(n) ⊑ σσ(n + 1) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞, ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 139 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], σσ(n) ⊑ σσ(n + 1) ⊦ ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ , ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [σσ(n), σσ(n + 1)]; [A] → [A], <all paths>

Additional information

Used terms: σσ(n), σσ(n + 1)

Proof step 140 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 141 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ ≠ [], ∃ σ1. σσ(n) + σ1 = σσ(n + 1) ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞, ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 142 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n) + σ1 = σσ(n + 1), σσ ≠ [] ⊦ ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ , ∃ n. # σσ(n) = ∞, ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: σ1

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 143 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n) + ⌜x⌝ = σσ(n + 1), σσ ≠ [] ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞, ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: x

Additional information

Used terms: x

Proof step 144 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n) + ⌜nf⌝ = σσ(n + 1), σσ ≠ [] ⊦ ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n))), ∃ n. # σσ(n) = ∞, ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 2 terms: n + 1 abort: F discard: T computed: F

Additional information

Used terms: n + 1

Proof step 145 from proof lim-exists (previous step) (back to tree)

# σσ = ∞, σσ(n) + ⌜nf⌝ = σσ(n + 1), # σσ(n + 1) ≠ ∞, σσ ≠ [] ⊦ ∃ x. σσ(n + 1) = σσ(n) + ⌜x⌝ , ∃ σ. (∀ n. σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. σ.take(m) ⊑ σσ(n)))

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 2, flags: rotate, discard

Additional information

Used simplifier rules: ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ n∞ + ∞ = ∞; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old

Proof step 146 from proof lim-exists (previous step) (back to tree)

# σσ ≠ ∞, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-inf, name: notinf-fin, seq: ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ ), subst:[n∞] → [# σσ], <all paths>

Additional information

Used terms: # σσ

Proof step 147 from proof lim-exists (previous step) (back to tree)

⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

A spec-lemma

Proof step 148 from proof lim-exists (previous step) (back to tree)

σσ ≠ [], mon(σσ), ∃ n. # σσ = ⌜n⌝ ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 149 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n⌝ , σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: n

Additional information

Used simplifier rules: Nat freely generated by 0, +1; ⊦ # σ = ⌜0⌝ ↔ σ = []; (s)

Proof step 150 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n⌝ , n = n0.+1, σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m.+1 = m + 1; (s)

Proof step 151 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ) ⊦ ∃ σ. (∀ n. ¬ n0 < n → σσ(n) ⊑ σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ σ.take(m) ⊑ σσ(n)))

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: σσ(n0) abort: F discard: T computed: F

Additional information

Used terms: σσ(n0)

Proof step 152 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ) ⊦ (∀ n. ¬ n0 < n → σσ(n) ⊑ σσ(n0)) ∧ (∀ m. ¬ # σσ(n0) < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ σσ(n0).take(m) ⊑ σσ(n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 153 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ) ⊦ (¬ n0 < n → σσ(n) ⊑ σσ(n0)) ∧ (¬ # σσ(n0) < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ σσ(n0).take(m) ⊑ σσ(n)))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 154 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), ¬ n0 < n, ¬ σσ(n) ⊑ σσ(n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 155 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), n ≤ n0, ¬ σσ(n) ⊑ σσ(n0) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:mon-lem, seq:mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n), subst:[σσ, m, n] → [σσ, n, n0]; [A] → [A]

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: σσ, n, n0

Proof step 156 from proof lim-exists (previous step) (back to tree)

mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n)

Lemma mon-lem

Proof step 157 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), ¬ # σσ(n0) < ⌜m⌝ ⊦ ∃ n. ¬ n0 < n ∧ σσ(n0).take(m) ⊑ σσ(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 158 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), ¬ # σσ(n0) < ⌜m⌝ ⊦ ∃ n. ¬ n0 < n ∧ σσ(n0).take(m) ⊑ σσ(n)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n0 abort: F discard: T computed: T

Additional information

Used terms: n0

Proof step 159 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), ¬ # σσ(n0) < ⌜m⌝ ⊦ ¬ n0 < n0 ∧ σσ(n0).take(m) ⊑ σσ(n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)

Proof step 160 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), ¬ # σσ(n0) < ⌜m⌝ , ¬ σσ(n0).take(m) ⊑ σσ(n0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream-old, name: prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [σσ(n0).take(m), σσ(n0)]; [A] → [A], <all paths>

Additional information

Used terms: σσ(n0).take(m), σσ(n0)

Proof step 161 from proof lim-exists (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 162 from proof lim-exists (previous step) (back to tree)

# σσ = ⌜n0⌝ + ⌜1⌝ , σσ ≠ [], mon(σσ), ¬ # σσ(n0) < ⌜m⌝ , ∀ σ1. σσ(n0).take(m) + σ1 ≠ σσ(n0) ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: σσ(n0).drop(m) abort: F discard: T computed: F

Additional information

Used simplifier rules: ¬ # σ < ⌜n⌝ ⊦ σ.take(n) + σ.drop(n) = σ ↔ true; (ss) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old Used terms: σσ(n0).drop(m)

Proof Tree for lim-exists (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree