Proof step 1 from proof nextli-lemma (conclusion) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 2:

  ∀ n. ∃ lf. lf(0) = [] ∧ (∀ m. m < n → lf(m + 1) = nextli(lf(m))) ⊦ ∃ lf. lf(0) = [] ∧ (∀ n. lf(n + 1) = nextli(lf(n)))

• Proof step 29:

  ⊦ ∀ n. ∃ lf. lf(0) = [] ∧ (∀ m. m < n → lf(m + 1) = nextli(lf(m))), ∃ lf. lf(0) = [] ∧ (∀ n. lf(n + 1) = nextli(lf(n)))

Rule argument

Additional information

Used formulas: ∀ n. ∃ lf. lf(0) = [] ∧ (∀ m. m < n → lf(m + 1) = nextli(lf(m)))

Proof step 2 from proof nextli-lemma (previous step) (back to tree)

User applied rule choice left

And obtained the following new goals

• Proof step 3:

  ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦ ∃ lf. lf(0) = [] ∧ (∀ n. lf(n + 1) = nextli(lf(n)))

Rule argument

Additional information

Proof step 3 from proof nextli-lemma (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 4:

  ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦ f_nat->nat->list<'a>(0)(0) = [] ∧ (∀ n. f_nat->nat->list<'a>(n + 1)(n + 1) = nextli(f_nat->nat->list<'a>(n)(n)))

Rule argument

Additional information

Used terms: λ n. f_nat->nat->list<'a>(n)(n)

Proof step 4 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 5:

  f_nat->nat->list<'a>(0)(0) = [], ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m. m < 0 → f_nat->nat->list<'a>(0)(m + 1) = nextli(f_nat->nat->list<'a>(0)(m)) ⊦ f_nat->nat->list<'a>(0)(0) = [] ∧ (∀ n. f_nat->nat->list<'a>(n + 1)(n + 1) = nextli(f_nat->nat->list<'a>(n)(n)))

Rule argument

Additional information

Used terms: 0

Proof step 5 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 6 from proof nextli-lemma (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 7:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m) ⊦

• Proof step 13:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦ ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m)

Rule argument

Additional information

Used formulas: ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m)

Proof step 7 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Proof step 8 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 9:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), n ≤ n ∧ n ≤ n + 1 → f_nat->nat->list<'a>(n)(n) = f_nat->nat->list<'a>(n + 1)(n), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m) ⊦

Rule argument

Additional information

Used terms: n, n, n + 1

Proof step 9 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n)(n) = f_nat->nat->list<'a>(n + 1)(n), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

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 ≤ m + n; (s) ⊦ m ≤ m; (s)

Proof step 10 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 11:

  f_nat->nat->list<'a>(n)(n) = f_nat->nat->list<'a>(n + 1)(n), f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n + 1)(0) = [], f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m), ∀ m. m < n + 1 → f_nat->nat->list<'a>(n + 1)(m + 1) = nextli(f_nat->nat->list<'a>(n + 1)(m)) ⊦

Rule argument

Additional information

Used terms: n + 1

Proof step 11 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  f_nat->nat->list<'a>(n + 1)(0) = [], f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n)(n) = f_nat->nat->list<'a>(n + 1)(n), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → f_nat->nat->list<'a>(n1)(m) = f_nat->nat->list<'a>(n2)(m), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m. ¬ n < m → f_nat->nat->list<'a>(n + 1)(m + 1) = nextli(f_nat->nat->list<'a>(n + 1)(m)) ⊦

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)

Proof step 12 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

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

Proof step 13 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m) ≠ f_nat->nat->list<'a>(n2)(m), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m ≤ n1, n1 ≤ n2, m ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat

Proof step 14 from proof nextli-lemma (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 15:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(0) ≠ f_nat->nat->list<'a>(n2)(0), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), 0 ≤ n2, n1 ≤ n2, 0 ≤ n1, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

• Proof step 19:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0 +1) ≠ f_nat->nat->list<'a>(n2)(m0 +1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 +1 ≤ n2, n1 ≤ n2, m0 +1 ≤ n1, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ f_nat->nat->list<'a>, n1, n2, n, nextli. ¬ ( f_nat->nat->list<'a>(0)(0) = [] ∧ f_nat->nat->list<'a>(n1)(m0) ≠ f_nat->nat->list<'a>(n2)(m0) ∧ f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)) ∧ m0 ≤ n1 ∧ n1 ≤ n2 ∧ m0 ≤ n2 ∧ (∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))))) ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 15 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(0) ≠ f_nat->nat->list<'a>(n2)(0), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), n1 ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)

Proof step 16 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 17:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(n1)(0) ≠ f_nat->nat->list<'a>(n2)(0), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), n1 ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m. m < n1 → f_nat->nat->list<'a>(n1)(m + 1) = nextli(f_nat->nat->list<'a>(n1)(m)) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 17 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(0)(0) = [], [] ≠ f_nat->nat->list<'a>(n2)(0), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), n1 ≤ n2, ∀ m. m < n1 → f_nat->nat->list<'a>(n1)(m + 1) = nextli(f_nat->nat->list<'a>(n1)(m)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Proof step 18 from proof nextli-lemma (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n2

Proof step 19 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n2, m0 < n1, n1 ≤ n2, ∀ f_nat->nat->list<'a>0, n0, n3, n4, nextli0. f_nat->nat->list<'a>0(0)(0) ≠ [] ∨ f_nat->nat->list<'a>0(n0)(m0) = f_nat->nat->list<'a>0(n3)(m0) ∨ f_nat->nat->list<'a>0(n4 + 1)(n4 + 1) = nextli0(f_nat->nat->list<'a>0(n4)(n4)) ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ n3 ∨ ¬ m0 ≤ n3 ∨ ¬ (∀ n0. f_nat->nat->list<'a>0(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>0(n0)(m + 1) = nextli0(f_nat->nat->list<'a>0(n0)(m)))), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

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

Proof step 20 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 21:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n1, m0 < n2, n1 ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), f_nat->nat->list<'a>(0)(0) ≠ [] ∨ f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0) ∨ f_nat->nat->list<'a>(n + 1)(n + 1) = nextli(f_nat->nat->list<'a>(n)(n)) ∨ ¬ m0 ≤ n1 ∨ ¬ n1 ≤ n2 ∨ ¬ m0 ≤ n2 ∨ ¬ (∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m)))) ⊦

Rule argument

Additional information

Used terms: f_nat->nat->list<'a>, n1, n2, n, nextli

Proof step 21 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n2, m0 < n1, n1 ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)

Proof step 22 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 23:

  f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n1, m0 < n2, n1 ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))), ∀ m. m < n1 → f_nat->nat->list<'a>(n1)(m + 1) = nextli(f_nat->nat->list<'a>(n1)(m)) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 23 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n2, m0 < n1, n1 ≤ n2, ∀ m. m < n1 → f_nat->nat->list<'a>(n1)(m + 1) = nextli(f_nat->nat->list<'a>(n1)(m)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Proof step 24 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 25:

  f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n1, m0 < n2, n1 ≤ n2, m0 < n1 → f_nat->nat->list<'a>(n1)(m0 + 1) = nextli(f_nat->nat->list<'a>(n1)(m0)), ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Used terms: m0

Proof step 25 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(n1)(m0 + 1) = nextli(f_nat->nat->list<'a>(n2)(m0)), f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n2, m0 < n1, n1 ≤ n2, ∀ n0. f_nat->nat->list<'a>(n0)(0) = [] ∧ (∀ m. m < n0 → f_nat->nat->list<'a>(n0)(m + 1) = nextli(f_nat->nat->list<'a>(n0)(m))) ⊦

Rule argument

Additional information

Proof step 26 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 27:

  f_nat->nat->list<'a>(n1)(m0 + 1) = nextli(f_nat->nat->list<'a>(n2)(m0)), f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(n2)(0) = [], f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n1, m0 < n2, n1 ≤ n2, ∀ m. m < n2 → f_nat->nat->list<'a>(n2)(m + 1) = nextli(f_nat->nat->list<'a>(n2)(m)) ⊦

Rule argument

Additional information

Used terms: n2

Proof step 27 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  f_nat->nat->list<'a>(n2)(0) = [], f_nat->nat->list<'a>(n1)(0) = [], f_nat->nat->list<'a>(0)(0) = [], f_nat->nat->list<'a>(n1)(m0) = f_nat->nat->list<'a>(n2)(m0), f_nat->nat->list<'a>(n1)(m0 + 1) = nextli(f_nat->nat->list<'a>(n2)(m0)), f_nat->nat->list<'a>(n1)(m0 + 1) ≠ f_nat->nat->list<'a>(n2)(m0 + 1), f_nat->nat->list<'a>(n + 1)(n + 1) ≠ nextli(f_nat->nat->list<'a>(n)(n)), m0 < n2, m0 < n1, n1 ≤ n2, ∀ m. m < n2 → f_nat->nat->list<'a>(n2)(m + 1) = nextli(f_nat->nat->list<'a>(n2)(m)) ⊦

Rule argument

Additional information

Proof step 28 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 29 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  ⊦ ∃ lf. lf(0) = [] ∧ (∀ n. lf(n + 1) = nextli(lf(n))), ∃ lf. lf(0) = [] ∧ (∀ m. m < n → lf(m + 1) = nextli(lf(m)))

Rule argument

Additional information

Proof step 30 from proof nextli-lemma (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 31:

  ⊦ ∃ lf. lf(0) = [] ∧ (∀ m. m < n → lf(m + 1) = nextli(lf(m)))

Rule argument

Additional information

Proof step 31 from proof nextli-lemma (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 32:

  ⊦ ∃ lf. lf(0) = [] ∧ (∀ m. m < 0 → lf(m + 1) = nextli(lf(m)))

• Proof step 34:

  ∀ nextli. ∃ lf. lf(0) = [] ∧ (∀ m. m < n0 → lf(m + 1) = nextli(lf(m))) ⊦ ∃ lf. lf(0) = [] ∧ (∀ m. m < n0 +1 → lf(m + 1) = nextli(lf(m)))

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 32 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  ⊦ ∃ lf. lf(0) = []

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 33 from proof nextli-lemma (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: λ n. []

Proof step 34 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  ∀ nextli. ∃ lf. lf(0) = [] ∧ (∀ m. m < n0 → lf(m + 1) = nextli(lf(m))) ⊦ ∃ lf. lf(0) = [] ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextli(lf(m)))

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)

Proof step 35 from proof nextli-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 36:

  ∃ lf. lf(0) = [] ∧ (∀ m. m < n0 → lf(m + 1) = nextli(lf(m))) ⊦ ∃ lf. lf(0) = [] ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextli(lf(m)))

Rule argument

Additional information

Used terms: nextli

Proof step 36 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  lf(0) = [], ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦ ∃ lf. lf(0) = [] ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextli(lf(m)))

Rule argument

Additional information

Proof step 37 from proof nextli-lemma (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 38:

  lf(0) = [], ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦ (0 = n0 + 1 ⊃ nextli(lf(n0));lf(0)) = [] ∧ (∀ m. ¬ n0 < m → (m + 1 = n0 + 1 ⊃ nextli(lf(n0));lf(m + 1)) = nextli((m = n0 + 1 ⊃ nextli(lf(n0));lf(m))))

Rule argument

Additional information

Used terms: λ n1. (n1 = n0 + 1 ⊃ nextli(lf(n0));lf(n1))

Proof step 38 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  lf(0) = [], (m = n0 ⊃ nextli(lf(n0));lf(m + 1)) ≠ nextli((m = n0 + 1 ⊃ nextli(lf(n0));lf(m))), m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦

Rule argument

Additional information

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

Proof step 39 from proof nextli-lemma (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 40:

  lf(0) = [], lf(m + 1) ≠ nextli((m = n0 + 1 ⊃ nextli(lf(n0));lf(m))), m ≠ n0, m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦

Rule argument

Additional information

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

Proof step 40 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  lf(0) = [], lf(m + 1) ≠ nextli((m = n0 + 1 ⊃ nextli(lf(n0));lf(m))), m ≠ n0, m < n0, ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦

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 ↔ m < n); (s)

Proof step 41 from proof nextli-lemma (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 42:

  lf(0) = [], lf(m + 1) ≠ nextli(lf(m)), m ≠ n0 + 1, m ≠ n0, m < n0, ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 42 from proof nextli-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

  lf(0) = [], lf(m + 1) ≠ nextli(lf(m)), m ≠ n0 + 1, m ≠ n0, m < n0, ∀ m. m < n0 → lf(m + 1) = nextli(lf(m)) ⊦

Rule argument

Additional information

Proof step 43 from proof nextli-lemma (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m