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: σ, σ0

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 73:

  # 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 43:

  # 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

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), ⌜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(n0) ⊦

Rule argument

Additional information

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

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + σ0 = ⌜x⌝ + σ1 ↔ σ0 = σ1; (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:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 32:

  # σ = # σ0, 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, ∃ σ3. ⌜σ0.take(n)⌝ + σ3 = ⌜σ.take(n)⌝ + σ2 + σ1 ⊦

Rule argument

Additional information

Used terms: ⌜σ0.take(n)⌝ , ⌜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)

Heuristic simplifier applied rule simplifier

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 ≠ [], 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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

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

• Proof step 35:

  #(⌜σ0.take(n)⌝ + σ3) = #(⌜σ.take(n)⌝ + σ2 + σ1), # σ = # σ0, 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, ∀ 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 34 from proof lim-unique (previous step) (back to tree)

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  # σ = # σ0, # σ3 = # σ2 + # σ1, 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, ∀ n2. ¬ ⌜n⌝ + # σ3 < ⌜n2⌝ → (⌜σ0.take(n)⌝ + σ3).take(n2) = (⌜σ.take(n)⌝ + σ2 + σ1).take(n2) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = n; (s) ⊦ # ⌜σ.take(n)⌝ = ⌜# σ.take(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 36 from proof lim-unique (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 37:

  # σ3 = # σ2 + # σ1, # σ = # σ0, σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, mon(σσ), ¬ # σ0 < ⌜n⌝ , ¬ ⌜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 37 from proof lim-unique (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  # 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)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 39:

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

• Proof step 40:

  # σ3 = # σ2 + # σ1, # σ = # σ0, ⌜σ0.take(n)⌝ .take(n) = (⌜σ.take(n)⌝ + σ2 + σ1).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, mon(σσ), ¬ # ⌜σ0.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 + σ0).take(n) = σ.l + σ0.take(n - # σ.l); (s) ⊦ ⌜x⌝ .l = x; (s) ⊦ m - m = 0; (s) ⊦ σ.take(0) = []; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜σ.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 39 from proof lim-unique (previous step) (back to tree)

A spec-lemma

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, # σ3 = # σ2 + # σ1, σ0.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: ⊦ m ≤ m; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = n; (s) ⊦ # ⌜σ.take(n)⌝ = ⌜# σ.take(n)⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ σ.take(n).take(n) = σ.take(n); (s) ⊦ ⌜x⌝ .take(n) = x.take(n); (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).take(n) = σ.take(n)

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ¬ # σ < ⌜n⌝ ⊦ σ.take(n).take(n) = σ.take(n); (s) ⊦ # σ < ⌜n⌝ → (σ + σ0).take(n) = σ.l + σ0.take(n - # σ.l); (s) ⊦ ⌜x⌝ .l = x; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = n; (s) ⊦ m - m = 0; (s) ⊦ σ.take(0) = []; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream Used terms: ⌜σ.take(n)⌝ , σ2 + σ1, n

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, 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

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 ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 46:

  # 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, ∃ σ1. ⌜σ0.take(n)⌝ + σ1 = σσ(n1) ⊦

Rule argument

Additional information

Used terms: ⌜σ0.take(n)⌝ , σσ(n1)

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, ⌜σ0.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), ⌜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 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 ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 49:

  # σ = # σ0, ⌜σ0.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. ⌜σ.take(n)⌝ + σ1 = σσ(n1) ⊦

Rule argument

Additional information

Used terms: ⌜σ.take(n)⌝ , σσ(n1)

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, ⌜σ.take(n)⌝ + σ2 = σσ(n1), ⌜σ0.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 50 from proof lim-unique (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 51:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 52:

  # σ = # σ0, ⌜σ0.take(n)⌝ + σ1 = σσ(n1), ⌜σ.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 51 from proof lim-unique (previous step) (back to tree)

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

  # σ = # σ0, ⌜σ0.take(n)⌝ + σ3 = σσ(n0), ⌜σ.take(n)⌝ + σ2 = σσ(n1), ⌜σ0.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 53 from proof lim-unique (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 54:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 55:

  # σ = # σ0, ⌜σ0.take(n)⌝ + σ1 = σσ(n1), ⌜σ.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 54 from proof lim-unique (previous step) (back to tree)

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

  # σ = # σ0, σσ(n0) + σ4 = σσ(n1), ⌜σ0.take(n)⌝ + σ3 = σσ(n0), ⌜σ.take(n)⌝ + σ2 = σσ(n1), ⌜σ0.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 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)⌝ + σ1 = ⌜σ.take(n)⌝ + σ2, ⌜σ0.take(n)⌝ + σ3 = σσ(n0), σσ(n0) + σ4 = ⌜σ.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 57 from proof lim-unique (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  # σ = # σ0, σσ(n0) + σ4 = ⌜σ.take(n)⌝ + σ2, ⌜σ0.take(n)⌝ + σ3 = σσ(n0), ⌜σ0.take(n)⌝ + σ1 = σσ(n0) + σ4, 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 58 from proof lim-unique (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 59:

  # σ = # σ0, ⌜σ0.take(n)⌝ + σ1 = (⌜σ0.take(n)⌝ + σ3) + σ4, (⌜σ0.take(n)⌝ + σ3) + σ4 = ⌜σ.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 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, (⌜σ0.take(n)⌝ + σ3) + σ4 = ⌜σ.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

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ ⌜x⌝ + σ0 = ⌜x⌝ + σ1 ↔ σ0 = σ1; (s)

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:

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

• Proof step 62:

  #((⌜σ0.take(n)⌝ + σ3) + σ4) = #(⌜σ.take(n)⌝ + σ2), # σ = # σ0, 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, ∀ n0. ¬ #((⌜σ0.take(n)⌝ + σ3) + σ4) < ⌜n0⌝ → ((⌜σ0.take(n)⌝ + σ3) + σ4).take(n0) = (⌜σ.take(n)⌝ + σ2).take(n0) ⊦

Rule argument

Additional information

Used terms: (⌜σ0.take(n)⌝ + σ3) + σ4, ⌜σ.take(n)⌝ + σ2

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:

  # σ = # σ0, # σ3 + # σ4 = # 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, ∀ n2. ¬ ⌜n⌝ + # σ2 < ⌜n2⌝ → ((⌜σ0.take(n)⌝ + σ3) + σ4).take(n2) = (⌜σ.take(n)⌝ + σ2).take(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.take(n) = n; (s) ⊦ # ⌜σ.take(n)⌝ = ⌜# σ.take(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 63 from proof lim-unique (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 64:

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

Rule argument

Additional information

Used terms: n

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, # σ3 + # σ4 = # σ2, ((⌜σ0.take(n)⌝ + σ3) + σ4).take(n) = (⌜σ.take(n)⌝ + σ2).take(n), 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

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 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)

• Proof step 67:

  # σ = # σ0, # σ3 + # σ4 = # σ2, (⌜σ0.take(n)⌝ + σ3).take(n) = (⌜σ.take(n)⌝ + σ2).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ #(⌜σ0.take(n)⌝ + σ3) < ⌜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 + σ0).take(n) = σ.l + σ0.take(n - # σ.l); (s) ⊦ #(σ + σ1) = # σ + # σ1; (s) ⊦ # ⌜σ.take(n)⌝ = ⌜# σ.take(n)⌝ ; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(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: ⌜σ0.take(n)⌝ + σ3, σ4, 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, # σ3 + # σ4 = # σ2, (⌜σ0.take(n)⌝ + σ3).take(n) = (⌜σ.take(n)⌝ + σ2).take(n), 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

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.take(n) = n; (s) ⊦ # ⌜σ.take(n)⌝ = ⌜# σ.take(n)⌝ ; (s) ⊦ #(σ + σ1) = # σ + # σ1; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 69:

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

• Proof step 70:

  # σ = # σ0, # σ3 + # σ4 = # σ2, ⌜σ0.take(n)⌝ .take(n) = (⌜σ.take(n)⌝ + σ2).take(n), σ.take(n) ≠ σ0.take(n), σσ ≠ [], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ¬ # ⌜σ0.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 + σ0).take(n) = σ.l + σ0.take(n - # σ.l); (s) ⊦ ⌜x⌝ .l = x; (s) ⊦ m - m = 0; (s) ⊦ σ.take(0) = []; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # ⌜σ.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 69 from proof lim-unique (previous step) (back to tree)

A spec-lemma

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:

  # σ = # σ0, # σ3 + # σ4 = # σ2, σ0.take(n) = (⌜σ.take(n)⌝ + σ2).take(n), 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

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 72:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ¬ # σ < ⌜n⌝ ⊦ σ.take(n).take(n) = σ.take(n); (s) ⊦ # σ < ⌜n⌝ → (σ + σ0).take(n) = σ.l + σ0.take(n - # σ.l); (s) ⊦ ⌜x⌝ .l = x; (s) ¬ # σ < ⌜n⌝ ⊦ # σ.take(n) = n; (s) ⊦ m - m = 0; (s) ⊦ σ.take(0) = []; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream Used terms: ⌜σ.take(n)⌝ , σ2, n

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 74:

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

User applied rule structural induction

And obtained the following new goals

• Proof step 75:

  ∞ = # 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 90:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 76:

  ∞ = # 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 76 from proof lim-unique (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 77:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 78:

  ∞ = # 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 77 from proof lim-unique (previous step) (back to tree)

A spec-lemma

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:

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

User applied rule all left

And obtained the following new goals

• Proof step 80:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 81:

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 82:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 83:

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 84:

  # 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 84 from proof lim-unique (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 85:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 86:

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 87:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 88:

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 89:

  ⌜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 ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 89 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.take(n)⌝ = ⌜# σ.take(n)⌝ ; (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 ⊦ ⌜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 90 from proof lim-unique (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 91:

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

User applied rule cut formula

And obtained the following new goals

• Proof step 92:

  ⌜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 103:

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

User applied rule all left

And obtained the following new goals

• Proof step 93:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 94:

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 95:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 96:

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 97:

  ⌜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 97 from proof lim-unique (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 98:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 99:

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 100:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 101:

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 102:

  ⌜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 ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 102 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.take(n)⌝ = ⌜# σ.take(n)⌝ ; (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 ⊦ ⌜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 103 from proof lim-unique (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 104:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 105:

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

User applied rule all left

And obtained the following new goals

• Proof step 106:

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 107:

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

User applied rule all left

And obtained the following new goals

• Proof step 108:

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

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 109:

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

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 110:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 111:

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

Rule argument

Additional information

Used terms: σ, σ0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 112:

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

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ σ1 ⊑ σ1 + σ2; (s) leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ #(σ + σ1) = # σ + # σ1; (s)

Proof step 112 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 leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ 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