Proof step 1 from proof max-ins-max (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  max(ins≤(a, x)) ≠ a, ¬ a < max(x) ⊦

Rule argument

Additional information

Proof step 2 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  max(ins≤(a, x)) ≠ a, ¬ a < max(x) ⊦

Rule argument

Additional information

Proof step 3 from proof max-ins-max (previous step) (back to tree)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

• Proof step 4:

  max(ins≤(a, a0 + x0)) ≠ a, ¬ a < max(a0 + x0), ∀ a. ¬(max(ins≤(a, x0)) ≠ a ∧ ¬ a < max(x0)) ⊦

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; list freely generated by [], +;; ⊦ ins≤(a, []) = a '; (s)

Proof step 4 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  max(ins≤(a, a0 ' + x0)) ≠ a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)

Proof step 5 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ ordered≤(a ' + x) → ins≤(a, x) = a ' + x

• Proof step 7:

  max(ins≤(a, a0 ' + x0)) ≠ a, ¬ a < max(a0 ' + x0), ¬ ordered≤(a ' + a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

• Proof step 37:

  max(a ' + a0 ' + x0) ≠ a, ordered≤(a ' + a0 ' + x0), ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used terms: a0 ' + x0, a

Proof step 6 from proof max-ins-max (previous step) (back to tree)

A spec-lemma

Proof step 7 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  max(ins≤(a, a0 ' + x0)) ≠ a, ¬ a < max(a0 ' + x0), ¬ ordered≤(a ' + a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Proof step 8 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b ' + x)

• Proof step 10:

  max(ins≤(a, a0 ' + x0)) ≠ a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), a0 < a ∨ ¬ ordered≤(a0 ' + x0) ⊦

Rule argument

Additional information

Used terms: a, a0, x0

Proof step 9 from proof max-ins-max (previous step) (back to tree)

A spec-lemma

Proof step 10 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  max(ins≤(a, a0 ' + x0)) ≠ a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), a0 < a ∨ ¬ ordered≤(a0 ' + x0) ⊦

Rule argument

Additional information

Proof step 11 from proof max-ins-max (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 12:

  max(ins≤(a, a0 ' + x0)) ≠ a, a0 < a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), a0 < a ∨ ¬ ordered≤(a0 ' + x0) ⊦

• Proof step 32:

  max(ins≤(a, a0 ' + x0)) ≠ a, ¬ a0 < a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), a0 < a ∨ ¬ ordered≤(a0 ' + x0) ⊦

Rule argument

Additional information

Used formulas: a0 < a

Proof step 12 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  max(a0 ' + ins≤(a, x0)) ≠ a, a0 < a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s) ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); (s)

Proof step 13 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

• Proof step 15:

  (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) ≠ a, ins≤(a, x0) ≠ [], a0 < a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ins≤(a, x) ≠ []; (s) Used terms: ins≤(a, x0), a0

Proof step 14 from proof max-ins-max (previous step) (back to tree)

Lemma max-rec

Proof step 15 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) ≠ a, a0 < a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ins≤(a, x) ≠ []; (s)

Proof step 16 from proof max-ins-max (previous step) (back to tree)

Heuristic cut applied rule cut formula

And obtained the following new goals

• Proof step 17:

  (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) ≠ a, a0 < a, max(ins≤(a, x0)) < a0, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

• Proof step 27:

  (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) ≠ a, a0 < a, ¬ max(ins≤(a, x0)) < a0, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used formulas: max(ins≤(a, x0)) < a0

Proof step 17 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  a0 ≠ a, max(ins≤(a, x0)) < a0, a0 < a, max(ins≤(a, x0)) < a, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem

Proof step 18 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ ordered≤(a ' + x) → ins≤(a, x) = a ' + x

• Proof step 20:

  a0 ≠ a, max(ins≤(a, x0)) < a, a0 < a, max(ins≤(a, x0)) < a0, ¬ a < max(a0 ' + x0), ¬ ordered≤(a ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: max-ge-first: ⊦ max(a ' + x) < a ↔ false; a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0, a

Proof step 19 from proof max-ins-max (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  a0 ≠ a, max(ins≤(a, x0)) < a0, a0 < a, max(ins≤(a, x0)) < a, ¬ a < max(a0 ' + x0), ¬ ordered≤(a ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Proof step 21 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ ordered≤(y) ↔ # y ≤ 1 ∨ ¬ y.tail.head < y.head ∧ ordered≤(y.tail)

• Proof step 23:

  a0 ≠ a, max(ins≤(a, x0)) < a, a0 < a, max(ins≤(a, x0)) < a0, ¬ a < max(a0 ' + x0), ¬ #(a ' + x0) ≤ 1, ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), (a ' + x0).tail.head < (a ' + x0).head ∨ ¬ ordered≤((a ' + x0).tail) ⊦

Rule argument

Additional information

Used terms: a ' + x0

Proof step 22 from proof max-ins-max (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  a0 ≠ a, x0 ≠ [], max(ins≤(a, x0)) < a0, a0 < a, max(ins≤(a, x0)) < a, ¬ a < (max(x0) < a0 ⊃ a0;max(x0)), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), x0.head < a ∨ ¬ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)); a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (a ' + x).tail = x; (s) ⊦ (a ' + x).head = a; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ # x = 0 ↔ x = []; (s) ⊦ m + m0 ≤ m ↔ m0 = 0; (s) ⊦ # a ' = 0 +1; (s) ⊦ #(x + y) = # x + # y; (s)

Proof step 24 from proof max-ins-max (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 25:

  x0 = a2 + x1, a0 ≠ a, x0 ≠ [], max(ins≤(a, x0)) < a, a0 < a, max(ins≤(a, x0)) < a0, ¬ a < (max(x0) < a0 ⊃ a0;max(x0)), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0), a2 < a ∨ ¬ ordered≤(x0) ⊦

Rule argument

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x0, a2, y

Proof step 25 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  a0 ≠ a, a2 < a, a0 < a, max(a2 ' + ins≤(a, x1)) < a0, max(a2 ' + ins≤(a, x1)) < a, ¬ a < (max(a2 ' + x1) < a0 ⊃ a0;max(a2 ' + x1)), ∀ a1. max(ins≤(a1, a2 ' + x1)) = a1 ∨ a1 < max(a2 ' + x1) ⊦

Rule argument

Additional information

Used simplifier rules: max-ge-first: ⊦ max(a ' + x) < a ↔ false; transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s) ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); (s) ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; (s) ⊦ a ' + x ≠ []; (s)

Proof step 26 from proof max-ins-max (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x); (s) ⊦ a + x = a ' + x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem ⊦ ¬ a < a; (s) transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem ⊦ a < b → (b < a ↔ false); (ss) Used terms: a

Proof step 27 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  max(ins≤(a, x0)) ≠ a, a0 < a, ¬ max(ins≤(a, x0)) < a0, ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Proof step 28 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

• Proof step 30:

  max(ins≤(a, x0)) ≠ a, x0 ≠ [], a0 < a, ¬ max(ins≤(a, x0)) < a0, ¬ a < (max(x0) < a0 ⊃ a0;max(x0)), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ ins≤(a, []) = a '; (s) Used terms: x0, a0

Proof step 29 from proof max-ins-max (previous step) (back to tree)

Lemma max-rec

Proof step 30 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  max(ins≤(a, x0)) ≠ a, x0 ≠ [], a0 < a, ¬ max(ins≤(a, x0)) < a0, ¬ a < (max(x0) < a0 ⊃ a0;max(x0)), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Proof step 31 from proof max-ins-max (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem ⊦ a < b → (b < a ↔ false); (ss) transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem ⊦ ¬ a < a; (s) Used terms: a

Proof step 32 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  max(a ' + a0 ' + x0) ≠ a, ¬ a0 < a, ¬ a < max(a0 ' + x0), ¬ a0 < max(a0 ' + x0), ¬ ordered≤(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s) ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; (s)

Proof step 33 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

  ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

• Proof step 35:

  a0 ' + x0 ≠ [], (max(a0 ' + x0) < a ⊃ a;max(a0 ' + x0)) ≠ a, ¬ a0 < a, ¬ a < max(a0 ' + x0), ¬ a0 < max(a0 ' + x0), ¬ ordered≤(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) ⊦ a ' ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a0 ' + x0, a

Proof step 34 from proof max-ins-max (previous step) (back to tree)

Lemma max-rec

Proof step 35 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  (max(a0 ' + x0) < a ⊃ a;max(a0 ' + x0)) ≠ a, ¬ a0 < a, ¬ a < max(a0 ' + x0), ¬ a0 < max(a0 ' + x0), ¬ ordered≤(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' + x ≠ []; (s)

Proof step 36 from proof max-ins-max (previous step) (back to tree)

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss) Used formulas: max(a0 ' + x0) < a

Proof step 37 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  max(a ' + a0 ' + x0) ≠ a, ordered≤(a ' + a0 ' + x0), ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Proof step 38 from proof max-ins-max (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 39:

  ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

• Proof step 40:

  a0 ' + x0 ≠ [], (max(a0 ' + x0) < a ⊃ a;max(a0 ' + x0)) ≠ a, ordered≤(a ' + a0 ' + x0), ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) ⊦ a ' ≠ []; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a0 ' + x0, a

Proof step 39 from proof max-ins-max (previous step) (back to tree)

Lemma max-rec

Proof step 40 from proof max-ins-max (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  (max(a0 ' + x0) < a ⊃ a;max(a0 ' + x0)) ≠ a, ordered≤(a ' + a0 ' + x0), ¬ a < max(a0 ' + x0), ∀ a1. max(ins≤(a1, x0)) = a1 ∨ a1 < max(x0) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' + x ≠ []; (s)

Proof step 41 from proof max-ins-max (previous step) (back to tree)

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss) Used formulas: max(a0 ' + x0) < a