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.take(m) ⊑ σσ(n)), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(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), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m) ⊑ σσ(n)) ⊦

Rule argument

Additional information

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.take(n) = σ1.take(n))

• Proof step 5:

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

• Proof step 67:

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

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

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

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.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, mon(σσ), σ0.take(n) ⊑ σσ(n0), ¬ # σ0 < ⌜n⌝ , ¬ # σ0 < ⌜n⌝ → (∃ n0. ⌜n0⌝ < # σσ ∧ σ.take(n) ⊑ σσ(n0)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

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.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), σ0.take(n) ⊑ σσ(n0), σ.take(n) ⊑ σσ(n1), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ 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.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), σ.take(n) ⊑ σσ(n1), σ0.take(n) ⊑ σσ(n0), ¬ # σ0 < ⌜n⌝ , ¬ n0 ≤ n1, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

• Proof step 42:

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

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)

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.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, mon(σσ), σ0.take(n) ⊑ σσ(n0), σ.take(n) ⊑ σσ(n1), σσ(n1) ⊑ σσ(n0), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (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.take(n) ≠ σ0.take(n), σσ ≠ [], n1 < n0, ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), σσ(n1) ⊑ σσ(n0), σ.take(n) ⊑ σσ(n1), σ0.take(n) ⊑ σσ(n0), σ.take(n) ⊑ σσ(n0), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream-old

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 20:

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

Rule argument

Additional information

Used terms: σσ(n1), σσ(n0)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 23:

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

Rule argument

Additional information

Used terms: σ.take(n), σσ(n1)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 26:

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

Rule argument

Additional information

Used terms: σ0.take(n), σσ(n0)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

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

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 31:

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

• Proof step 32:

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

Rule argument

Additional information

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

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

A spec-lemma

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

User applied rule all left

And obtained the following new goals

• Proof step 33:

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

Rule argument

Additional information

Used terms: n

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, # σ3 = # σ2 + # σ1, (σ0.take(n) + σ3).take(n) = ((σ.take(n) + σ2) + σ1).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], n1 < n0, ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ 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 ⊦ ¬ n∞ + n∞1 < n∞1; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

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

• Proof step 36:

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

Rule argument

Additional information

Used simplifier rules: ⊦ #(σ + σ1) = # σ + # σ1; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ¬ 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: σ.take(n) + σ2, σ1, n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  # σ = # σ0, # σ3 = # σ2 + # σ1, (σ0.take(n) + σ3).take(n) = (σ.take(n) + σ2).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], n1 < n0, ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ 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 ⊦ ¬ n∞ + n∞1 < n∞1; (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 38:

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

• Proof step 39:

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

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 41:

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

Rule argument

Additional information

Used simplifier rules: ¬ # σ < ⌜n⌝ ⊦ σ.take(n).take(n) = σ.take(n); (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ¬ n < n; (s) Used terms: σ0.take(n), σ3, n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

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

Rule argument

Additional information

Used simplifier rules: leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream-old

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 44:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 45:

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

Rule argument

Additional information

Used terms: σ.take(n), σσ(n1)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 47:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 48:

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

Rule argument

Additional information

Used terms: σ0.take(n), σσ(n1)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 50:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 51:

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

Rule argument

Additional information

Used terms: σ0.take(n), σσ(n0)

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

A spec-lemma

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

Rule argument

Additional information

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:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 54:

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

Rule argument

Additional information

Used terms: σσ(n0), σσ(n1)

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:

  # σ = # σ0, σσ(n0) + σ4 = σσ(n1), σ0.take(n) + σ3 = σσ(n0), σ0.take(n) + σ2 = σσ(n1), σ.take(n) + σ1 = σσ(n1), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0 ⊦

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 56:

  # σ = # σ0, σ.take(n) + σ1 = σ0.take(n) + σ2, σ0.take(n) + σ3 = σσ(n0), σσ(n0) + σ4 = σ0.take(n) + σ2, σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 57:

  # σ = # σ0, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, σ.take(n) + σ1 = σ0.take(n) + σ2, σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0 ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 58:

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

• Proof step 59:

  #(σ.take(n) + σ1) = #(σ0.take(n) + σ2), # σ = # σ0, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n0. ¬ #(σ.take(n) + σ1) < ⌜n0⌝ → (σ.take(n) + σ1).take(n0) = (σ0.take(n) + σ2).take(n0) ⊦

Rule argument

Additional information

Used terms: σ.take(n) + σ1, σ0.take(n) + σ2

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 60:

  # σ = # σ0, # σ1 = # σ2, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n2. ¬ ⌜n⌝ + # σ2 < ⌜n2⌝ → (σ.take(n) + σ1).take(n2) = (σ0.take(n) + σ2).take(n2) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = ⌜n⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (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 60 from proof lim-unique (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 61:

  # σ1 = # σ2, # σ = # σ0, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ¬ ⌜n⌝ + # σ2 < ⌜n⌝ → (σ.take(n) + σ1).take(n) = (σ0.take(n) + σ2).take(n), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 62:

  # σ = # σ0, # σ1 = # σ2, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, (σ.take(n) + σ1).take(n) = (σ0.take(n) + σ2).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ 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 62 from proof lim-unique (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 63:

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

• Proof step 64:

  # σ1 = # σ2, # σ = # σ0, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, (σ.take(n) + σ1).take(n) = σ0.take(n).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ¬ # σ0.take(n) < ⌜n⌝ , ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  # σ = # σ0, # σ1 = # σ2, (σ0.take(n) + σ3) + σ4 = σ0.take(n) + σ2, (σ.take(n) + σ1).take(n) = σ0.take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ # σ0 < ⌜n⌝ , ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0 ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 66:

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

Rule argument

Additional information

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

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

A spec-lemma

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:

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

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 69:

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

• Proof step 84:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 70:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 71:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 72:

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

Rule argument

Additional information

Used terms: # σ0

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

A spec-lemma

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:

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

Rule argument

Additional information

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

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

Rule argument

Additional information

Used terms: n + 1

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:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 76:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 77:

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

Rule argument

Additional information

Used terms: σ.take(n + 1), σσ(n0)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 78:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 79:

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

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 80:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 81:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 82:

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

Rule argument

Additional information

Used terms: σσ(n0), σ0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 83:

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

Rule argument

Additional information

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

Proof step 83 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) ⊦ ¬ ∞ < 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 append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ ⌜n⌝ + n∞ = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n = 0; (ws) 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 84 from proof lim-unique (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 85:

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

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 86:

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

• Proof step 97:

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

Rule argument

Additional information

Used formulas: ⌜m⌝ < # σ0

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

User applied rule all left

And obtained the following new goals

• Proof step 87:

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

Rule argument

Additional information

Used terms: m + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 88:

  ⌜m⌝ = # σ, σσ ≠ [], ⌜m⌝ ≠ # σ0, ⌜m⌝ < # σ0, ⌜n⌝ < # σσ, mon(σσ), σ0.take(m + 1) ⊑ σσ(n), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ 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 88 from proof lim-unique (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 89:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 90:

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

Rule argument

Additional information

Used terms: σ0.take(m + 1), σσ(n)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 91:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 92:

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

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 93:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 94:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 95:

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

Rule argument

Additional information

Used terms: σσ(n), σ

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 96:

  σ0.take(m + 1) + σ1 = σσ(n), ⌜m⌝ = # σσ(n) + # σ2, σσ ≠ [], ⌜m⌝ ≠ # σ0, ⌜n⌝ < # σσ, ⌜m⌝ < # σ0, mon(σσ), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m0. ¬ m < m0 → (∃ n0. ⌜n0⌝ < # σσ ∧ (σσ(n) + σ2).take(m0) ⊑ σσ(n0)) ⊦

Rule argument

Additional information

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

Proof step 96 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.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 append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream-old ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ; (s) ⊦ ⌜n⌝ + n∞ = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n = 0; (ws) 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 97 from proof lim-unique (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 98:

  ⌜m⌝ = # σ, ⌜m0⌝ = # σ0, σσ ≠ [], ⌜m⌝ ≠ # σ0, mon(σσ), ¬ ⌜m⌝ < # σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m) ⊑ σσ(n)) ⊦

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 99:

  ⌜m0⌝ = # σ0, ⌜m⌝ = # σ, m ≠ m0, σσ ≠ [], m0 < m, mon(σσ), ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

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) ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss) ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 100:

  ⌜m⌝ = # σ, ⌜m0⌝ = # σ0, m ≠ m0, σσ ≠ [], m0 < m, mon(σσ), ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ σ.take(m) ⊑ σσ(n)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)) ⊦

Rule argument

Additional information

Used terms: m

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 101:

  ⌜m⌝ = # σ, ⌜m0⌝ = # σ0, m ≠ m0, σσ ≠ [], m0 < m, ⌜n⌝ < # σσ, mon(σσ), σ ⊑ σσ(n), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # σ = ⌜n⌝ → σ.take(n) = σ; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 102:

  ⊦ σ1 ⊑ σ2 → (σ2 ⊑ σ1 ↔ σ1 = σ2)

• Proof step 103:

  ⌜m0⌝ = # σ0, ⌜m⌝ = # σ, m ≠ m0, σσ ≠ [], ⌜n⌝ < # σσ, m0 < m, mon(σσ), σ ⊑ σσ(n), ¬ σσ(n) ⊑ σ, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

• Proof step 105:

  σσ(n) = σ, ⌜m0⌝ = # σ0, ⌜m⌝ = # σ, m ≠ m0, σσ ≠ [], ⌜n⌝ < # σσ, m0 < m, mon(σσ), σσ(n) ⊑ σ, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ ⊦

Rule argument

Additional information

Used terms: σ, σσ(n)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 104:

  ⌜m⌝ = # σ, ⌜m0⌝ = # σ0, m ≠ m0, σ ≠ σσ(n), σσ ≠ [], m0 < m, ⌜n⌝ < # σσ, mon(σσ), σ ⊑ σσ(n), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ, ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)) ⊦

Rule argument

Additional information

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

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 106:

  ⌜m⌝ = # σσ(n), ⌜m0⌝ = # σ0, m ≠ m0, σσ ≠ [], m0 < m, ⌜n⌝ < # σσ, mon(σσ), ∀ n0. ⌜n0⌝ < # σσ → σσ(n0) ⊑ σσ(n), ∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ σ ⊑ σ; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 107:

  ⌜m0⌝ = # σ0, ⌜m⌝ = # σσ(n), m ≠ m0, σσ ≠ [], ⌜n⌝ < # σσ, m0 < m, mon(σσ), ⌜n⌝ < # σσ → σσ(n) ⊑ σ0, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)), ∀ n0. ⌜n0⌝ < # σσ → σσ(n0) ⊑ σσ(n) ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 108:

  ⌜m⌝ = # σσ(n), ⌜m0⌝ = # σ0, m ≠ m0, σσ ≠ [], m0 < m, ⌜n⌝ < # σσ, mon(σσ), σσ(n) ⊑ σ0, ∀ n0. ⌜n0⌝ < # σσ → σσ(n0) ⊑ σσ(n), ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 109:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 110:

  ⌜m0⌝ = # σ0, ⌜m⌝ = # σσ(n), m ≠ m0, σσ ≠ [], ⌜n⌝ < # σσ, m0 < m, mon(σσ), ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ σ0.take(m1) ⊑ σσ(n)), ∀ n0. ⌜n0⌝ < # σσ → σσ(n0) ⊑ σσ(n), ∃ σ1. σσ(n) + σ1 = σ0 ⊦

Rule argument

Additional information

Used terms: σσ(n), σ0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 111:

  ⌜m⌝ = # σσ(n), ⌜m0⌝ = ⌜m⌝ + # σ1, m ≠ m0, σσ ≠ [], m0 < m, ⌜n⌝ < # σσ, mon(σσ), ∀ n0. ⌜n0⌝ < # σσ → σσ(n0) ⊑ σσ(n), ∀ m1. ¬ m0 < m1 → (∃ n0. ⌜n0⌝ < # σσ ∧ (σσ(n) + σ1).take(m1) ⊑ σσ(n0)) ⊦

Rule argument

Additional information

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

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

User applied rule constructor cut

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (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 ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m = m + n ↔ n = 0; (s) ⊦ # x = 0 ↔ x = []; (s) ⊦ m + m0 < n → ¬ n + n0 < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic