Proof step 1 from proof leninv-zero (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(leninv(nσ) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝ ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 2 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ¬(leninv(nσ) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝ ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 3 from proof leninv-zero (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ leninv(nσ) = max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ(m0) = n - 1))))

• Proof step 5:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ(m0) = n - 1)))) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof leninv-zero (previous step) (back to tree)

Lemma leninv-def

Proof step 5 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝

Rule argument

Additional information

Proof step 6 from proof leninv-zero (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 8:

  stutf(nσ) ⊦ (0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = 0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = 0 - 1)))) ∧ (∀ m. (m ≠ 0 → (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)))) → ¬ 0 < m) ↔ nσ = ⌜λ n. 0⌝

Rule argument

Additional information

Used terms: 0, λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))

Proof step 7 from proof leninv-zero (previous step) (back to tree)

Lemma maxinf-fin

Proof step 8 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  nσ ≠ ⌜λ n. 0⌝ , stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1))) → m = 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → n ≠ 0; (s) ⊦ 0 < m ↔ m ≠ 0; (s)

Proof step 9 from proof leninv-zero (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1))) → m = 0, # nσ ≠ # ⌜λ n. 0⌝ ∨ (∃ n. ¬(⌜n⌝ < # nσ → nσ(n) = ⌜λ n. 0⌝ (n))) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. 0⌝ , nσ

Proof step 10 from proof leninv-zero (previous step) (back to tree)

A spec-lemma

Proof step 11 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1))) → m = 0, # nσ ≠ ∞ ∨ ¬(⌜n⌝ < # nσ → nσ(n) = 0) ⊦

Rule argument

Additional information

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

Proof step 12 from proof leninv-zero (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1))) → m = 0, # nσ ≠ ∞ ∨ ¬(⌜n⌝ < # nσ → nσ(n) = 0) ⊦

• Proof step 19:

  # nσ ≠ ∞, stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1))) → m = 0, # nσ ≠ ∞ ∨ ¬(⌜n⌝ < # nσ → nσ(n) = 0) ⊦

Rule argument

Additional information

Used formulas: # nσ = ∞

Proof step 13 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # nσ = ∞, nσ(n) ≠ 0, stutf(nσ), ∀ m. (∃ m0. nσ(m0) = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)) → m = 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 14 from proof leninv-zero (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 15:

  # nσ = ∞, nσ(n) ≠ 0, stutf(nσ), ∀ m. (∃ m0. nσ(m0) = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)) → m = 0, (∃ m0. nσ(m0) = 0.+1 - 1 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = 0.+1 - 1)) → 0.+1 = 0 ⊦

Rule argument

Additional information

Used terms: 0.+1

Proof step 15 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # nσ = ∞, nσ(n) ≠ 0, stutf(nσ), ∀ m. (∃ m0. nσ(m0) = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)) → m = 0 ⊦ ∃ m0. nσ(m0) = 0 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = 0)

Rule argument

Additional information

Proof step 16 from proof leninv-zero (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 17:

  # nσ = ∞, nσ(n) ≠ 0, stutf(nσ), ∀ m. (∃ m0. nσ(m0) = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)) → m = 0 ⊦ ∃ m0. nσ(m0) = 0 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = 0), nσ(0) = 0 ∧ ¬(∀ m1. 0 < m1 → nσ(m1) = 0)

Rule argument

Additional information

Used terms: 0

Proof step 17 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # nσ = ∞, nσ(n) ≠ 0, stutf(nσ), ∀ m. (∃ m0. nσ(m0) = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = m - 1)) → m = 0, ∀ m1. m1 ≠ 0 → nσ(m1) = 0 ⊦ ∃ m0. nσ(m0) = 0 ∧ ¬(∀ m1. m0 < m1 → nσ(m1) = 0)

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ(0) = 0; ⊦ 0 < m ↔ m ≠ 0; (s)

Proof step 18 from proof leninv-zero (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ(0) = 0; Used terms: n

Proof step 19 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ ≠ ∞, stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1) → m = 0 ⊦

Rule argument

Additional information

Proof step 20 from proof leninv-zero (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 22:

  stutf(nσ), ∀ m. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ(m0) = m - 1) → m = 0, ∃ n. # nσ = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # nσ

Proof step 21 from proof leninv-zero (previous step) (back to tree)

A spec-lemma

Proof step 22 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # nσ = ⌜n⌝ , stutf(nσ), ∀ m. (∃ m0. m0 < n ∧ nσ(m0) = m - 1) → m = 0 ⊦

Rule argument

Additional information

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

Proof step 23 from proof leninv-zero (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 24:

  # nσ = ⌜n⌝ , n = n0.+1, stutf(nσ), ∀ m. (∃ m0. m0 < n ∧ nσ(m0) = m - 1) → m = 0 ⊦

Rule argument

Additional information

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

Proof step 24 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ(m0) = m - 1) → m = 0 ⊦

Rule argument

Additional information

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

Proof step 25 from proof leninv-zero (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 26:

  # nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ(m0) = m - 1) → m = 0, (∃ m0. ¬ n0 < m0 ∧ nσ(m0) = nσ(n0) + 1 - 1) → nσ(n0) + 1 = 0 ⊦

Rule argument

Additional information

Used terms: nσ(n0) + 1

Proof step 26 from proof leninv-zero (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ(m0) = m - 1) → m = 0 ⊦ ∃ m0. ¬ n0 < m0 ∧ nσ(m0) = nσ(n0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≠ 0; (s) ⊦ n + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)

Proof step 27 from proof leninv-zero (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: n0