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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  σ ≠ σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ0 to m ⊑ σσ(n)), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ to m ⊑ σσ(n)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  σ ≠ σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ σ1 ⊑ σ2 ↔ (∃ σ. σ1 + σ = σ2); (ls)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

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

• Proof step 5:

  σσ ≠ [], mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), # σ ≠ # σ0 ∨ (∃ n. ¬(¬ # σ < ⌜n⌝ → σ to n = σ0 to n)) ⊦

Rule argument

Additional information

Used terms: σ0, σ

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  σσ ≠ [], mon(σσ), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), # σ ≠ # σ0 ∨ ¬(¬ # σ < ⌜n⌝ → σ to n = σ0 to n) ⊦

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 7:

  # σ = # σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), # σ ≠ # σ0 ∨ ¬(¬ # σ < ⌜n⌝ → σ to n = σ0 to n) ⊦

• Proof step 49:

  # σ ≠ # σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), # σ ≠ # σ0 ∨ ¬(¬ # σ < ⌜n⌝ → σ to n = σ0 to n) ⊦

Rule argument

Additional information

Used formulas: # σ = # σ0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # σ = # σ0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 9:

  # σ = # σ0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ¬ # σ0 < ⌜n⌝ → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ0. (σ to n) + σ0 = σσ(n0))) ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 11:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ¬ # σ0 < ⌜n⌝ → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ. (σ0 to n) + σ = σσ(n0))) ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 13:

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

• Proof step 14:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ¬ n0 ≤ n1, ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

• Proof step 33:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, σσ(n0) ⊑ σσ(n1), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used terms: σσ, n0, n1

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

Lemma mon-lem

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, n1 < n0, ⌜n1⌝ < # σσ, σσ(n1) ⊑ σσ(n0), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → ¬ n < m; (s) Used terms: σσ, n1, n0

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

Lemma mon-lem

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), σσ(n1) + σ3 = σσ(n0), n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ σ1 ⊑ σ2 ↔ (∃ σ. σ1 + σ = σ2); (ls)

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

User applied rule subst equation

And obtained the following new goals

• Proof step 19:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), ((σ0 to n) + σ2) + σ3 = σσ(n0), n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, n1 < n0, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 20:

  # σ = # σ0, ((σ0 to n) + σ2) + σ3 = (σ to n) + σ1, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

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

• Proof step 22:

  #(((σ0 to n) + σ2) + σ3) = #((σ to n) + σ1), # σ = # σ0, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, n1 < n0, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n0. ¬ #(((σ0 to n) + σ2) + σ3) < ⌜n0⌝ → ((σ0 to n) + σ2) + σ3 to n0 = (σ to n) + σ1 to n0 ⊦

Rule argument

Additional information

Used terms: (σ to n) + σ1, ((σ0 to n) + σ2) + σ3

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # σ = # σ0, # σ2 + # σ3 = # σ1, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n2. ¬ ⌜n⌝ + # σ1 < ⌜n2⌝ → ((σ0 to n) + σ2) + σ3 to n2 = (σ to n) + σ1 to n2 ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to 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 ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 24:

  # σ2 + # σ3 = # σ1, # σ = # σ0, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, n1 < n0, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ¬ ⌜n⌝ + # σ1 < ⌜n⌝ → ((σ0 to n) + σ2) + σ3 to n = (σ to n) + σ1 to n ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # σ = # σ0, # σ2 + # σ3 = # σ1, ((σ0 to n) + σ2) + σ3 to n = (σ to n) + σ1 to n, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

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∞ + n∞1 < n∞1; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ ¬ # σ < ⌜n⌝ → σ + σ0 to n = σ to n

• Proof step 27:

  # σ2 + # σ3 = # σ1, # σ = # σ0, (σ0 to n) + σ2 to n = (σ to n) + σ1 to n, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, n1 < n0, ⌜n1⌝ < # σσ, ¬ #((σ0 to n) + σ2) < ⌜n⌝ , ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ #(σ + σ1) = # σ + # σ1; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ¬ n∞ + n∞1 < n∞1; (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: n, σ3, (σ0 to n) + σ2

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  # σ = # σ0, # σ2 + # σ3 = # σ1, (σ0 to n) + σ2 to n = (σ to n) + σ1 to n, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

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∞ + n∞1 < n∞1; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ ¬ # σ < ⌜n⌝ → σ + σ0 to n = σ to n

• Proof step 30:

  # σ2 + # σ3 = # σ1, # σ = # σ0, σ0 to n to n = (σ to n) + σ1 to n, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, n1 < n0, ⌜n1⌝ < # σσ, ¬ #(σ0 to n) < ⌜n⌝ , ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < n; (s) Used terms: n, σ2, σ0 to n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  # σ = # σ0, # σ2 + # σ3 = # σ1, σ0 to n = (σ to n) + σ1 to n, n0 ≠ 0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ σ to n to n = σ to n; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

  ⊦ ¬ # σ < ⌜n⌝ → σ + σ0 to n = σ to n

Rule argument

Additional information

Used simplifier rules: ¬ # σ < ⌜n⌝ ⊦ σ to n to n = σ to n; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < n; (s) Used terms: n, σ1, σ to n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  # σ = # σ0, (σ to n) + σ1 = σσ(n0), (σ0 to n) + σ2 = σσ(n1), σσ(n0) + σ3 = σσ(n1), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ σ1 ⊑ σ2 ↔ (∃ σ. σ1 + σ = σ2); (ls)

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

User applied rule subst equation

And obtained the following new goals

• Proof step 35:

  # σ = # σ0, ((σ to n) + σ1) + σ3 = σσ(n1), (σ0 to n) + σ2 = σσ(n1), σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 36:

  # σ = # σ0, ((σ to n) + σ1) + σ3 = (σ0 to n) + σ2, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 37:

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

• Proof step 38:

  #(((σ to n) + σ1) + σ3) = #((σ0 to n) + σ2), # σ = # σ0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n0. ¬ #(((σ to n) + σ1) + σ3) < ⌜n0⌝ → ((σ to n) + σ1) + σ3 to n0 = (σ0 to n) + σ2 to n0 ⊦

Rule argument

Additional information

Used terms: (σ0 to n) + σ2, ((σ to n) + σ1) + σ3

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  # σ = # σ0, # σ1 + # σ3 = # σ2, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n2. ¬ ⌜n⌝ + # σ2 < ⌜n2⌝ → ((σ to n) + σ1) + σ3 to n2 = (σ0 to n) + σ2 to n2 ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to 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 ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 40:

  # σ1 + # σ3 = # σ2, # σ = # σ0, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ¬ ⌜n⌝ + # σ2 < ⌜n⌝ → ((σ to n) + σ1) + σ3 to n = (σ0 to n) + σ2 to n ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  # σ = # σ0, # σ1 + # σ3 = # σ2, ((σ to n) + σ1) + σ3 to n = (σ0 to n) + σ2 to n, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

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∞ + n∞1 < n∞1; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

  ⊦ ¬ # σ < ⌜n⌝ → σ + σ0 to n = σ to n

• Proof step 43:

  # σ1 + # σ3 = # σ2, # σ = # σ0, (σ to n) + σ1 to n = (σ0 to n) + σ2 to n, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ #((σ to n) + σ1) < ⌜n⌝ , ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ #(σ + σ1) = # σ + # σ1; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ¬ n∞ + n∞1 < n∞1; (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: n, σ3, (σ to n) + σ1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # σ = # σ0, # σ1 + # σ3 = # σ2, (σ to n) + σ1 to n = (σ0 to n) + σ2 to n, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

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∞ + n∞1 < n∞1; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 45:

  ⊦ ¬ # σ < ⌜n⌝ → σ + σ0 to n = σ to n

• Proof step 46:

  # σ1 + # σ3 = # σ2, # σ = # σ0, σ to n to n = (σ0 to n) + σ2 to n, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ #(σ to n) < ⌜n⌝ , ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < n; (s) Used terms: n, σ1, σ to n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  # σ = # σ0, # σ1 + # σ3 = # σ2, σ to n = (σ0 to n) + σ2 to n, σ to n ≠ σ0 to n, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ σ to n to n = σ to n; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 48:

  ⊦ ¬ # σ < ⌜n⌝ → σ + σ0 to n = σ to n

Rule argument

Additional information

Used simplifier rules: ¬ # σ < ⌜n⌝ ⊦ σ to n to n = σ to n; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < n; (s) Used terms: n, σ2, σ0 to n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  # σ ≠ # σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 51:

  ∞ = # σ, # σ ≠ # σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

• Proof step 62:

  ⌜n⌝ = # σ, # σ ≠ # σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: Natinf freely generated by ∞, ⌝;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 52:

  ∞ = # σ, σσ ≠ [], ∞ ≠ # σ0, mon(σσ), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n)), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 53:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 54:

  ∞ = # σ, σσ ≠ [], mon(σσ), ∀ m. ∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n)), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∃ n. # σ0 = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # σ0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 55:

  # σ0 = ⌜n⌝ , ∞ = # σ, σσ ≠ [], mon(σσ), ∀ m. ∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n)), ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 56:

  # σ0 = ⌜n⌝ , ∞ = # σ, σσ ≠ [], mon(σσ), ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ0. (σ to n + 1) + σ0 = σσ(n0)) ⊦

Rule argument

Additional information

Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  # σ0 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ(n0), ∞ = # σ, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 58:

  # σ0 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ(n0), ∞ = # σ, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ⌜n0⌝ < # σσ → (∃ σ. σσ(n0) + σ = σ0) ⊦

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  # σσ(n0) + # σ2 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ(n0), ∞ = # σ, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n0) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s)

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

User applied rule subst equation

And obtained the following new goals

• Proof step 60:

  #((σ to n + 1) + σ1) + # σ2 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ(n0), ∞ = # σ, σσ ≠ [], mon(σσ), ⌜n0⌝ < # σσ, ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n0) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 61:

  ⊦ #(σ + σ1) = # σ + # σ1

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ∞ < n∞; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to 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 ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ n∞ + n∞1 = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n∞1 = ⌜0⌝ ; (ws) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = 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 Used terms: σ1, σ to n + 1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  ⌜n⌝ = # σ, ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 64:

  ⌜n⌝ = # σ, ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ⌜n⌝ < # σ0, ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

• Proof step 69:

  ⌜n⌝ = # σ, ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ¬ ⌜n⌝ < # σ0, ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

Rule argument

Additional information

Used formulas: ⌜n⌝ < # σ0

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

User applied rule all left

And obtained the following new goals

• Proof step 65:

  ⌜n⌝ = # σ, ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ⌜n⌝ < # σ0, ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ¬ # σ0 < ⌜n + 1⌝ → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ. (σ0 to n + 1) + σ = σσ(n0))) ⊦

Rule argument

Additional information

Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 66:

  ⌜n⌝ = # σ, (σ0 to n + 1) + σ1 = σσ(n0), ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ⌜n⌝ < # σ0, ⌜n0⌝ < # σσ, ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ) ⊦

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 ⊦ # σ < ⌜n⌝ + ⌜1⌝ ↔ ¬ ⌜n⌝ < # σ; (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)

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

User applied rule all left

And obtained the following new goals

• Proof step 67:

  ⌜n⌝ = # σ, (σ0 to n + 1) + σ1 = σσ(n0), ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ⌜n⌝ < # σ0, ⌜n0⌝ < # σσ, ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ⌜n0⌝ < # σσ → (∃ σ0. σσ(n0) + σ0 = σ) ⊦

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 68:

  ⌜n⌝ = # σσ(n0) + # σ2, (σ0 to n + 1) + σ1 = σσ(n0), ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ⌜n⌝ < # σ0, ⌜n0⌝ < # σσ, ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σσ(n0) + σ2 to m) + σ0 = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)

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

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # σ < ⌜n⌝ + ⌜1⌝ ↔ ¬ ⌜n⌝ < # σ; (s) ¬ # σ < ⌜n⌝ ⊦ #(σ to 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 append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ n∞ + n∞1 = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n∞1 = ⌜0⌝ ; (ws) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = 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

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

User applied rule structural induction

And obtained the following new goals

• Proof step 70:

  ⌜n⌝ = # σ, ⌜n0⌝ = # σ0, ⌜n⌝ ≠ # σ0, σσ ≠ [], mon(σσ), ¬ ⌜n⌝ < # σ0, ∀ m. ¬ ⌜n⌝ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: Natinf freely generated by ∞, ⌝; ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ < ∞; (s)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 71:

  ⌜n⌝ = # σ, ⌜n0⌝ = # σ0, n ≠ n0, n ≠ 0, σσ ≠ [], mon(σσ), n0 < n, ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ0. (σ to m) + σ0 = σσ(n))), ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 72:

  ⌜n⌝ = # σ, ⌜n0⌝ = # σ0, n ≠ n0, n ≠ 0, σσ ≠ [], mon(σσ), n0 < n, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ¬ n < n → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ0. (σ to n) + σ0 = σσ(n0))) ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 73:

  ⌜n⌝ = # σ, ⌜n0⌝ = # σ0, σ + σ1 = σσ(n1), n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ. σσ(n) + σ = σ0), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)) ⊦

Rule argument

Additional information

Used simplifier rules: mon-def: ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1)); ⊦ # σ = ⌜n⌝ → σ to n = σ; (s) ⊦ ¬ n < n; (s) ⊦ σ1 ⊑ σ2 ↔ (∃ σ. σ1 + σ = σ2); (ls) ⊦ m < n → n ≠ 0; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 74:

  ⌜n⌝ = # σ, ⌜n0⌝ = # σ0, σ + σ1 = σσ(n1), n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σ0 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)), ⌜n1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σ0) ⊦

Rule argument

Additional information

Used terms: n1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 75:

  ⌜n⌝ = # σ, ⌜n0⌝ = # σσ(n1) + # σ2, σ + σ1 = σσ(n1), n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n1) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s)

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

User applied rule subst equation

And obtained the following new goals

• Proof step 76:

  ⌜n⌝ = # σ, ⌜n0⌝ = #(σ + σ1) + # σ2, n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n1) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 77:

  ⌜n⌝ = # σ, ⌜n0⌝ = (⌜n⌝ + # σ1) + # σ2, n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n1) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)) ⊦

Rule argument

Additional information

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

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

User applied rule structural induction

And obtained the following new goals

• Proof step 78:

  ⌜n⌝ = # σ, ⌜n0⌝ = (⌜n⌝ + # σ1) + # σ2, ⌜n2⌝ = # σ2, n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n1) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)) ⊦

Rule argument

Additional information

Used simplifier rules: Natinf freely generated by ∞, ⌝; ⊦ ∞ + ⌜m⌝ = ∞; (s) ⊦ 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)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 79:

  ⌜n⌝ = # σ, ⌜n0⌝ = (⌜n⌝ + # σ1) + ⌜n2⌝ , ⌜n2⌝ = # σ2, n ≠ n0, σσ ≠ [], n0 < n, ⌜n1⌝ < # σσ, ∀ m. ¬ n0 < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ. (σσ(n1) + σ2 to m) + σ = σσ(n))), ∀ n. ⌜n⌝ < # σσ → (∃ σ0. σσ(n) + σ0 = σ), ∀ n1. ⌜n1⌝ + ⌜1⌝ < # σσ → (∃ σ. σσ(n1) + σ = σσ(n1 + 1)) ⊦

Rule argument

Additional information

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

User applied rule structural induction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Natinf freely generated by ∞, ⌝; ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n1 + 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 = m + n ↔ n = 0; (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 ⊦ ∞ + ⌜m⌝ = ∞; (s) ⊦ ⌜n⌝ ≠ ∞; (s)