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Proof step 1 from proof monl-equiv (conclusion) (back to tree)

⊦ mon(lf) ↔ (∀ af. mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ))

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:monl-def, seq: ⊦ mon(lf) ↔ (∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2)), subst:[lf] → [lf]; [A, B] → [A, B], lhs:mon(lf), <all paths>

Additional information

Used terms: lf

Proof step 2 from proof monl-equiv (previous step) (back to tree)

⊦ mon(lf) ↔ (∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2))

Lemma monl-def

Proof step 3 from proof monl-equiv (previous step) (back to tree)

⊦ (∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2)) ↔ (∀ af. mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 4 from proof monl-equiv (previous step) (back to tree)

∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦ ∀ af. mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 5 from proof monl-equiv (previous step) (back to tree)

¬ mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ), ∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: streamseq-mon-lim, name:mon-def, seq: ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1)), subst:[σσ] → [⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ]; [A] → [B], lhs:mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ), <all paths>

Additional information

Used terms: ⌜λ n. ⌜lf(af.take(n))⌝ ⌝

Proof step 6 from proof monl-equiv (previous step) (back to tree)

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

A spec-lemma

Proof step 7 from proof monl-equiv (previous step) (back to tree)

∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2), ∃ n. ¬(⌜n⌝ + ⌜1⌝ < # ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ → ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n) ⊑ ⌜λ n. ⌜lf(af.take(n))⌝ ⌝ (n + 1)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(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 ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 8 from proof monl-equiv (previous step) (back to tree)

¬ ⌜lf(af.take(n))⌝ ⊑ ⌜lf(af.take(n + 1))⌝ , ∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 2 terms: af.take(n), af.take(n + 1) abort: F discard: T computed: F

Additional information

Used terms: af.take(n), af.take(n + 1)

Proof step 9 from proof monl-equiv (previous step) (back to tree)

¬ ⌜lf(af.take(n))⌝ ⊑ ⌜lf(af.take(n + 1))⌝ , af.take(n) ⊑ af.take(n + 1) → lf(af.take(n)) ⊑ lf(af.take(n + 1)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream, name:to-unwind-end-inf, seq: ⊦ nf.take(m + 1) = nf.take(m) + nf(m) ', subst:[nf, m] → [af, n]; [A] → [A], lhs:af.take(n + 1), <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: af, n

Proof step 10 from proof monl-equiv (previous step) (back to tree)

⊦ nf.take(m + 1) = nf.take(m) + nf(m) '

A spec-lemma

Proof step 11 from proof monl-equiv (previous step) (back to tree)

¬ ⌜lf(af.take(n))⌝ ⊑ ⌜lf(af.take(n) + af(n) ')⌝ , af.take(n) ⊑ af.take(n) + af(n) ' → lf(af.take(n)) ⊑ lf(af.take(n) + af(n) ') ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s)

Proof step 12 from proof monl-equiv (previous step) (back to tree)

lf(af.take(n)) ⊑ lf(af.take(n) + af(n) '), ¬ ⌜lf(af.take(n))⌝ ⊑ ⌜lf(af.take(n) + af(n) ')⌝ ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-last, name:prefix, seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y), subst:[x, y] → [lf(af.take(n)), lf(af.take(n) + af(n) ')]; [A] → [B], lhs:lf(af.take(n)) ⊑ lf(af.take(n) + af(n) '), <all paths>

Additional information

Used simplifier rules: ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ σ1 ⊑ σ1 + σ2; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream Used terms: lf(af.take(n)), lf(af.take(n) + af(n) ')

Proof step 13 from proof monl-equiv (previous step) (back to tree)

⊦ x ⊑ y ↔ (∃ z. x + z = y)

A spec-lemma

Proof step 14 from proof monl-equiv (previous step) (back to tree)

∀ af. mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) ⊦ ∀ al1, al2. al1 ⊑ al2 → lf(al1) ⊑ lf(al2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 15 from proof monl-equiv (previous step) (back to tree)

al1 ⊑ al2, ¬ lf(al1) ⊑ lf(al2), ∀ af. mon(⌜λ n. ⌜lf(af.take(n))⌝ ⌝ ) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: λ n. (n < # al2 ⊃ al2(n);a) abort: F discard: T computed: F

Additional information

Used terms: λ n. (n < # al2 ⊃ al2(n);a)

Proof step 16 from proof monl-equiv (previous step) (back to tree)

mon(⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, ¬ lf(al1) ⊑ lf(al2) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: streamseq-mon-lim, name:mon-lem, seq:mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ(m) ⊑ σσ(n), subst:[σσ, m, n] → [⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ , # al1, # al2]; [A] → [B]

Additional information

Used simplifier rules: ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ # y < # x → ¬ x ⊑ y; (s) Used terms: ⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ , # al1, # al2

Proof step 17 from proof monl-equiv (previous step) (back to tree)

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

A spec-lemma

Proof step 18 from proof monl-equiv (previous step) (back to tree)

mon(⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, ⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ (# al1) ⊑ ⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ (# al2), ¬ lf(al1) ⊑ lf(al2) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 19 from proof monl-equiv (previous step) (back to tree)

mon(⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), al1 ⊑ al2, ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(# al1))⌝ ⊑ ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(# al2))⌝ , ¬ lf(al1) ⊑ lf(al2) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-last, name:prefix, seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y), subst:[x, y] → [al1, al2]; [A] → [A], lhs:al1 ⊑ al2, <all paths>

Additional information

Used terms: al1, al2

Proof step 20 from proof monl-equiv (previous step) (back to tree)

⊦ x ⊑ y ↔ (∃ z. x + z = y)

A spec-lemma

Proof step 21 from proof monl-equiv (previous step) (back to tree)

mon(⌜λ n. ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(# al1))⌝ ⊑ ⌜lf((λ n. (n < # al2 ⊃ al2(n);a)).take(# al2))⌝ , ¬ lf(al1) ⊑ lf(al2), ∃ z. al1 + z = al2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ #(x + y) = # x + # y; (s)

Proof step 22 from proof monl-equiv (previous step) (back to tree)

mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1))⌝ ⊑ ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al))⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al

Additional information

Used formulas: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al

Proof step 23 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1))⌝ ⊑ ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al))⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 24 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1))⌝ ⊑ ⌜lf(al1 + al)⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1

Additional information

Used formulas: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1

Proof step 25 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1))⌝ ⊑ ⌜lf(al1 + al)⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 26 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf(al1)⌝ ⊑ ⌜lf(al1 + al)⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream, name:prefix-def, seq: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0), subst:[σ, σ0] → [⌜lf(al1)⌝ , ⌜lf(al1 + al)⌝ ]; [A] → [B], lhs:⌜lf(al1)⌝ ⊑ ⌜lf(al1 + al)⌝ , <all paths>

Additional information

Used terms: ⌜lf(al1)⌝ , ⌜lf(al1 + al)⌝

Proof step 27 from proof monl-equiv (previous step) (back to tree)

⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

A spec-lemma

Proof step 28 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ¬ lf(al1) ⊑ lf(al1 + al), ∃ β. ⌜lf(al1)⌝ + β = ⌜lf(al1 + al)⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 29 from proof monl-equiv (previous step) (back to tree)

⌜lf(al1)⌝ + β = ⌜lf(al1 + al)⌝ , (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ¬ lf(al1) ⊑ lf(al1 + al) ⊦

User applied rule constructor cut

And closed the goal!

Rule argument

var: β

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ ⌜x⌝ ≠ σ + ⌜nf0⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z; (s) ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 30 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) ≠ al1, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1))⌝ ⊑ ⌜lf(al1 + al)⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

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] → [(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1), al1]; [A] → [A], lhs:(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1) = al1, <all paths>

Additional information

Used simplifier rules: ⊦ # nf.take(n) = n; (s) ⊦ m < n → (n ≤ m ↔ false); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ n < # x → (x + y)(n) = x(n); (s) a: ⊦ (x + y) + z = x + y + z; from specification list f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ # x = 0 ↔ x = []; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1), al1

Proof step 31 from proof monl-equiv (previous step) (back to tree)

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

A spec-lemma

Proof step 32 from proof monl-equiv (previous step) (back to tree)

(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) ≠ al1 + al, mon(⌜λ n. ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(n))⌝ ⌝ ), ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1))⌝ ⊑ ⌜lf((λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al))⌝ , ¬ lf(al1) ⊑ lf(al1 + al) ⊦

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:[x, y] → [(λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al), al1 + al]; [A] → [A], lhs: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al) = al1 + al, <all paths>

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ # nf.take(n) = n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) Used terms: (λ n. (n < # al1 + # al ⊃ (al1 + al)(n);a)).take(# al1 + # al), al1 + al

Proof step 33 from proof monl-equiv (previous step) (back to tree)

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

A spec-lemma

Proof Tree for monl-equiv (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