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

User applied rule cut formula

And obtained the following new goals

• Proof step 2:

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

• Proof step 31:

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

Rule argument

Additional information

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

Proof step 2 from proof nextl-lemma-butlast (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. lf(0) = a ' ∧ (∀ n. lf(n + 1) = nextl(n, lf(n).last))

Rule argument

Additional information

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

User applied rule choice left

And obtained the following new goals

• Proof step 4:

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

Rule argument

Additional information

Proof step 4 from proof nextl-lemma-butlast (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(0)(0) = a ' ∧ (∀ n. lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last))

Rule argument

Additional information

Used terms: λ n. lfgen(n)(n)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

Proof step 6 from proof nextl-lemma-butlast (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)), ∀ m. m < 0 → lfgen(0)(m + 1) = nextl(m, lfgen(0)(m).last) ⊦ lfgen(0)(0) = a ' ∧ lfgen(n + 1)(n + 1) = nextl(n, lfgen(n)(n).last)

Rule argument

Additional information

Used terms: 0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

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

User applied rule cut formula

And obtained the following new goals

• Proof step 9:

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

• Proof step 15:

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

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 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 11:

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

Rule argument

Additional information

Used terms: n, n, n + 1

Proof step 11 from proof nextl-lemma-butlast (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), lfgen(n + 1)(n + 1) ≠ nextl(n, lfgen(n)(n).last), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last)) ⊦

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

Rule argument

Additional information

Used terms: n + 1

Proof step 13 from proof nextl-lemma-butlast (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), lfgen(n + 1)(n + 1) ≠ nextl(n, lfgen(n)(n).last), ∀ m, n1, n2. m ≤ n1 ∧ n1 ≤ n2 → lfgen(n1)(m) = lfgen(n2)(m), ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last)), ∀ m. ¬ n < m → lfgen(n + 1)(m + 1) = nextl(m, lfgen(n + 1)(m).last) ⊦

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

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 (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 17:

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

• Proof step 21:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

Used terms: n2

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

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

Rule argument

Additional information

Proof step 20 from proof nextl-lemma-butlast (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 21 from proof nextl-lemma-butlast (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  lfgen(0)(0) = a ', lfgen(n1)(m0 + 1) ≠ lfgen(n2)(m0 + 1), lfgen(n + 1)(n + 1) ≠ nextl(n, lfgen(n)(n).last), m0 < n2, m0 < n1, n1 ≤ n2, ∀ lfgen0, a0, n0, n3, n4, nextl0. lfgen0(0)(0) ≠ a0 ' ∨ lfgen0(n0)(m0) = lfgen0(n3)(m0) ∨ lfgen0(n4 + 1)(n4 + 1) = nextl0(n4, lfgen0(n4)(n4).last) ∨ ¬ m0 ≤ n0 ∨ ¬ n0 ≤ n3 ∨ ¬ m0 ≤ n3 ∨ ¬(∀ n0. lfgen0(n0)(0) = a0 ' ∧ (∀ m. m < n0 → lfgen0(n0)(m + 1) = nextl0(m, lfgen0(n0)(m).last))), ∀ n0. lfgen(n0)(0) = a ' ∧ (∀ m. m < n0 → lfgen(n0)(m + 1) = nextl(m, lfgen(n0)(m).last)) ⊦

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

User applied rule all left

And obtained the following new goals

• Proof step 23:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 25:

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

Rule argument

Additional information

Used terms: n1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 27:

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

Rule argument

Additional information

Used terms: m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 29:

  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(n + 1)(n + 1) ≠ nextl(n, lfgen(n)(n).last), m0 < n1, m0 < n2, n1 ≤ n2, ∀ m. m < n2 → lfgen(n2)(m + 1) = nextl(m, lfgen(n2)(m).last) ⊦

Rule argument

Additional information

Used terms: n2

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  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(n + 1)(n + 1) ≠ nextl(n, lfgen(n)(n).last), m0 < n2, m0 < n1, n1 ≤ n2, ∀ m. m < n2 → lfgen(n2)(m + 1) = nextl(m, lfgen(n2)(m).last) ⊦

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

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

Rule argument

Additional information

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

User applied rule weakening

And obtained the following new goals

• Proof step 33:

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

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 34:

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

• Proof step 36:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  ⊦ ∃ lf. lf(0) = a '

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 35 from proof nextl-lemma-butlast (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 36 from proof nextl-lemma-butlast (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

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

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

User applied rule all left

And obtained the following new goals

• Proof step 38:

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

Rule argument

Additional information

Used terms: a, nextl

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

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

Rule argument

Additional information

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

User applied rule exists right

And obtained the following new goals

• Proof step 40:

  lf(0) = a ', ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ⊦ (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))

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  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) ⊦

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 41 from proof nextl-lemma-butlast (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) = a ', lf(m + 1) ≠ nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last), m ≠ n0, m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ⊦

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 44:

  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) ⊦

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 44 from proof nextl-lemma-butlast (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, lf(m).last), m ≠ n0 + 1, m ≠ n0, m < n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ⊦

Rule argument

Additional information

Proof step 45 from proof nextl-lemma-butlast (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m