Proof step 1 from proof nextl-lemma-butlast-nonempty (conclusion) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 2:

  ∀ n, a. nextl(n, a) ≠ [], ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ []) ⊦ ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [])

• Proof step 32:

  ∀ n, a. nextl(n, a) ≠ [] ⊦ ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ []), ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [])

Rule argument

Additional information

Used formulas: ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

Proof step 2 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ∀ n. ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [])

Rule argument

Additional information

Proof step 3 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule choice left

And obtained the following new goals

• Proof step 4:

  ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦ ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [])

Rule argument

Additional information

Proof step 4 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 5:

  ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦ lfgen(0)(0) = a ' ∧ (∀ n. lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ [])

Rule argument

Additional information

Used terms: λ n. lfgen(n)(n)

Proof step 5 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦ lfgen(0)(0) = a ' ∧ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Proof step 6 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 7:

  lfgen(0)(0) = a ', ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < 0 → lfgen(0)(m + 1) = nextl(m, lfgen(0)(m).last) ∧ lfgen(0)(m) ≠ [] ⊦ lfgen(0)(0) = a ' ∧ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Used terms: 0

Proof step 7 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  lfgen(0)(0) = a ', ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 8 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 9:

  lfgen(0)(0) = a ', ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [], ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

• Proof step 15:

  lfgen(0)(0) = a ', ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦ ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Used formulas: ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m)

Proof step 9 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  lfgen(0)(0) = a ', ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Proof step 10 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 11:

  lfgen(0)(0) = a ', n ≤ n ∧ n ≤ n + 1 → lfgen(n)(n) = lfgen(n + 1)(n), ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [], ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Used terms: n, n, n + 1

Proof step 11 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  lfgen(0)(0) = a ', lfgen(n)(n) = lfgen(n + 1)(n), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

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 12 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 13:

  lfgen(n)(n) = lfgen(n + 1)(n), lfgen(0)(0) = a ', lfgen(n + 1)(0) = a ', ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [], ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ m. m < n + 1 → lfgen(n + 1)(m + 1) = nextl(m, lfgen(n + 1)(m).last) ∧ lfgen(n + 1)(m) ≠ [] ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

Used terms: n + 1

Proof step 13 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  lfgen(n + 1)(0) = a ', lfgen(0)(0) = a ', lfgen(n)(n) = lfgen(n + 1)(n), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ m. ¬ n < m → lfgen(n + 1)(m + 1) = nextl(m, lfgen(n + 1)(m).last) ∧ lfgen(n + 1)(m) ≠ [] ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

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 14 from proof nextl-lemma-butlast-nonempty (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 15 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  lfgen(0)(0) = a ', lfgen(n1)(m) ≠ lfgen(n2)(m), m ≤ n1, n1 ≤ n2, m ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last) ∧ lfgen(n)(n) ≠ []

Rule argument

Additional information

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

Proof step 16 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 17:

  lfgen(0)(0) = a ', lfgen(n1)(m) ≠ lfgen(n2)(m), m ≤ n2, n1 ≤ n2, m ≤ n1, ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Proof step 17 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 18:

  lfgen(0)(0) = a ', lfgen(n1)(0) ≠ lfgen(n2)(0), 0 ≤ n1, n1 ≤ n2, 0 ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦

• Proof step 22:

  lfgen(0)(0) = a ', lfgen(n1)(m0 +1) ≠ lfgen(n2)(m0 +1), m0 +1 ≤ n1, n1 ≤ n2, m0 +1 ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ lfgen, a, n1, n2, nextl. ¬ ( lfgen(0)(0) = a ' ∧ lfgen(n1)(m0) ≠ lfgen(n2)(m0) ∧ m0 ≤ n2 ∧ n1 ≤ n2 ∧ m0 ≤ n1 ∧ (∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ [])) ∧ (∀ n, a. nextl(n, a) ≠ [])) ⊦

Rule argument

Additional information

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

Proof step 18 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  lfgen(0)(0) = a ', lfgen(n1)(0) ≠ lfgen(n2)(0), n1 ≤ n2, ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)

Proof step 19 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 20:

  lfgen(0)(0) = a ', lfgen(n2)(0) = a ', lfgen(n1)(0) ≠ lfgen(n2)(0), n1 ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ m. m < n2 → lfgen(n2)(m + 1) = nextl(m, lfgen(n2)(m).last) ∧ lfgen(n2)(m) ≠ [] ⊦

Rule argument

Additional information

Used terms: n2

Proof step 20 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  lfgen(n2)(0) = a ', lfgen(0)(0) = a ', lfgen(n1)(0) ≠ a ', n1 ≤ n2, ∀ m. m < n2 → lfgen(n2)(m + 1) = nextl(m, lfgen(n2)(m).last) ∧ lfgen(n2)(m) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Proof step 21 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n1

Proof step 22 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  lfgen(0)(0) = a ', lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), m0 < n1, m0 < n2, n1 ≤ n2, ∀ lfgen0, a0, n, n0, nextl0. lfgen0(0)(0) ≠ a0 ' ∨ lfgen0(n)(m0) = lfgen0(n0)(m0) ∨ ¬ m0 ≤ n0 ∨ ¬ n ≤ n0 ∨ ¬ m0 ≤ n ∨ ¬(∀ n0. lfgen0(n0)(0) = a0 ' ∧ (∀ m. m < n0 → lfgen0(n0)(m + 1) = nextl0(m, lfgen0(n0)(m).last) ∧ lfgen0(n0)(m) ≠ [])) ∨ ¬(∀ n, a. nextl0(n, a) ≠ []), ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat 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 23 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 24:

  lfgen(0)(0) = a ', lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), m0 < n2, m0 < n1, n1 ≤ n2, ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [], lfgen(0)(0) ≠ a ' ∨ lfgen(n1)(m0) = lfgen(n2)(m0) ∨ ¬ m0 ≤ n2 ∨ ¬ n1 ≤ n2 ∨ ¬ m0 ≤ n1 ∨ ¬(∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ [])) ∨ ¬(∀ n, a. nextl(n, a) ≠ []) ⊦

Rule argument

Additional information

Used terms: lfgen, a, n1, n2, nextl

Proof step 24 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  lfgen(0)(0) = a ', lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), m0 < n1, m0 < n2, n1 ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦

Rule argument

Additional information

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

Proof step 25 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 26:

  lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(0)(0) = a ', lfgen(n1)(0) = a ', lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), m0 < n2, m0 < n1, n1 ≤ n2, ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n1 → lfgen(n1)(m + 1) = nextl(m, lfgen(n1)(m).last) ∧ lfgen(n1)(m) ≠ [] ⊦

Rule argument

Additional information

Used terms: n1

Proof step 26 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  lfgen(n1)(0) = a ', lfgen(0)(0) = a ', lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), m0 < n1, m0 < n2, n1 ≤ n2, ∀ m. m < n1 → lfgen(n1)(m + 1) = nextl(m, lfgen(n1)(m).last) ∧ lfgen(n1)(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦

Rule argument

Additional information

Proof step 27 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 28:

  lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(0)(0) = a ', lfgen(n1)(0) = a ', lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), m0 < n2, m0 < n1, n1 ≤ n2, m0 < n1 → lfgen(n1)(m0 + 1) = nextl(m0, lfgen(n1)(m0).last) ∧ lfgen(n1)(m0) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []), ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Used terms: m0

Proof step 28 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  lfgen(n1)(0) = a ', lfgen(0)(0) = a ', lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(n1)(m0 + 1) = nextl(m0, lfgen(n2)(m0).last), lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), lfgen(n2)(m0) ≠ [], m0 < n1, m0 < n2, n1 ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last) ∧ lfgen(n0)(m) ≠ []) ⊦

Rule argument

Additional information

Proof step 29 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 30:

  lfgen(n1)(m0 + 1) = nextl(m0, lfgen(n2)(m0).last), lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(0)(0) = a ', lfgen(n1)(0) = a ', lfgen(n2)(0) = a ', lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), lfgen(n2)(m0) ≠ [], m0 < n2, m0 < n1, n1 ≤ n2, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n2 → lfgen(n2)(m + 1) = nextl(m, lfgen(n2)(m).last) ∧ lfgen(n2)(m) ≠ [] ⊦

Rule argument

Additional information

Used terms: n2

Proof step 30 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  lfgen(n2)(0) = a ', lfgen(n1)(0) = a ', lfgen(0)(0) = a ', lfgen(n1)(m0) = lfgen(n2)(m0), lfgen(n1)(m0 + 1) = nextl(m0, lfgen(n2)(m0).last), lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), lfgen(n2)(m0) ≠ [], m0 < n1, m0 < n2, n1 ≤ n2, ∀ m. m < n2 → lfgen(n2)(m + 1) = nextl(m, lfgen(n2)(m).last) ∧ lfgen(n2)(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Proof step 31 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 32 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ []), ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

Rule argument

Additional information

Proof step 33 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 34:

  ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. m < n → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

Rule argument

Additional information

Proof step 34 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 35:

  ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. m < 0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

• Proof step 37:

  ∀ n, a. nextl(n, a) ≠ [], ∀ nextl, a. (∀ n, a. nextl(n, a) ≠ []) → (∃ lf. lf(0) = a ' ∧ (∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])) ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. m < n0 +1 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

Rule argument

Additional information

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

Proof step 35 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a '

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 36 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: λ n. a '

Proof step 37 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  ∀ nextl, a. (∀ n, a. nextl(n, a) ≠ []) → (∃ lf. lf(0) = a ' ∧ (∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])), ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextl(m, lf(m).last) ∧ 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 38 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 39:

  (∀ n, a. nextl(n, a) ≠ []) → (∃ lf. lf(0) = a ' ∧ (∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])), ∀ n, a. nextl(n, a) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

Rule argument

Additional information

Used terms: nextl, a

Proof step 39 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  lf(0) = a ', ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [])

Rule argument

Additional information

Proof step 40 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 41:

  lf(0) = a ', ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦ (0 = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(0)) = a ' ∧ (∀ m. ¬ n0 < m → (m + 1 = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last) ∧ (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) ≠ [])

Rule argument

Additional information

Used terms: λ n1. (n1 = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(n1))

Proof step 41 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ (m = n0 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last) ∧ (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);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 42 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 43:

  lf(0) = a ', (m = n0 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) ≠ nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last), m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

• Proof step 48:

  (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) = [], lf(0) = a ', m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Proof step 43 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 44:

  lf(0) = a ', lf(m + 1) ≠ nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last), m ≠ n0, m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ 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 44 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 45:

  lf(0) = a ', lf(m + 1) ≠ nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last), m ≠ n0, m < n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ 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 45 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 46:

  lf(0) = a ', lf(m + 1) ≠ nextl(m, lf(m).last), m ≠ n0 + 1, m ≠ n0, m < n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

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 46 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  lf(0) = a ', lf(m + 1) ≠ nextl(m, lf(m).last), m ≠ n0 + 1, m ≠ n0, m < n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦

Rule argument

Additional information

Proof step 47 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m

Proof step 48 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

  (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) = [], lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦

Rule argument

Additional information

Proof step 49 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 50:

  lf(0) = a ', lf(m) = [], m ≠ n0 + 1, m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

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

Proof step 50 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 51:

  lf(m) = [], lf(0) = a ', m ≠ n0 + 1, m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦

Rule argument

Additional information

Proof step 51 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 52:

  lf(0) = a ', lf(m) = [], m ≠ n0 + 1, m ≤ n0, m - 1 < n0 → lf((m - 1) + 1) = nextl(m - 1, lf(m - 1).last) ∧ lf(m - 1) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Used terms: m - 1

Proof step 52 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 53:

  lf(m) = [], lf(0) = a ', m = m0 + 1, m ≠ 0, m ≠ n0 + 1, m ≤ n0, m0 < n0 → lf(m0 + 1) = nextl(m0, lf(m0).last) ∧ lf(m0) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ 0 ≤ m; (s) ⊦ a ' ≠ []; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0

Proof step 53 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 54:

  [] = nextl(m0, lf(m0).last), lf(m0 + 1) = [], lf(0) = a ', lf(m0) ≠ [], m0 ≠ n0, m0 < n0, ∀ n, a. nextl(n, a) ≠ [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) 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 + n = m + n0 ↔ n = n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 54 from proof nextl-lemma-butlast-nonempty (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0, lf(m0).last