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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  max(x + x0) = max(x), ¬(max(x + x1) = max(x + x0) ↔ true), x ≠ [], ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  max(x + x0) = max(x), max(x + x1) ≠ max(x), x ≠ [], ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  max((a + x2) + x0) = max(a + x2), a + x2 ≠ [], max((a + x2) + x1) ≠ max(a + x2), ordered≤(x0), x1 ⊑ x0, ∀ x0, x1. ¬(max(x2 + x0) = max(x2) ∧ max(x2 + x1) ≠ max(x2) ∧ x2 ≠ [] ∧ ordered≤(x0) ∧ x1 ⊑ x0) ⊦

Rule argument

Additional information

Used simplifier rules: list freely generated by [], +;;

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  max((a ' + x2) + x0) = max(a ' + x2), max((a ' + x2) + x1) ≠ max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

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

• Proof step 7:

  max((a ' + x2) + x0) = max(a ' + x2), x2 + x1 ≠ [], (max(x2 + x1) < a ⊃ a;max(x2 + x1)) ≠ max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) ⊦ [] + x = x; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x1, a

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

Lemma max-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  max((a ' + x2) + x0) = max(a ' + x2), x2 + x1 ≠ [], (max(x2 + x1) < a ⊃ a;max(x2 + x1)) ≠ max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 9:

  max((a ' + x2) + x0) = max(a ' + x2), x2 + x1 ≠ [], a ≠ max(a ' + x2), max(x2 + x1) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

• Proof step 16:

  max((a ' + x2) + x0) = max(a ' + x2), x2 + x1 ≠ [], max(x2 + x1) ≠ max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(x2)), x2 + x1 ≠ [], a ≠ (max(x2) < a ⊃ a;max(x2)), x2 ≠ [], max(x2 + x1) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ [] + x = x; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2, a

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

Lemma max-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  max((a ' + x2) + x0) = max(x2), x2 + x1 ≠ [], a ≠ (max(x2) < a ⊃ a;max(x2)), x2 ≠ [], max(x2 + x1) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

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

• Proof step 14:

  (max(x2 + x0) < a ⊃ a;max(x2 + x0)) = max(x2), x2 + x0 ≠ [], x2 + x1 ≠ [], a ≠ (max(x2) < a ⊃ a;max(x2)), x2 ≠ [], max(x2 + x1) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x0, a

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

Lemma max-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  (max(x2 + x0) < a ⊃ a;max(x2 + x0)) = max(x2), x2 + x0 ≠ [], x2 + x1 ≠ [], a ≠ (max(x2) < a ⊃ a;max(x2)), x2 ≠ [], max(x2 + x1) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss) transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem a: ⊦ (x + y) + z = x + y + z; from specification list

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  max((a ' + x2) + x0) = max(a ' + x2), x2 + x1 ≠ [], max(x2 + x1) ≠ max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

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

• Proof step 19:

  max((a ' + x2) + x0) = max(a ' + x2), x2 = [], x2 + x1 ≠ [], max(x2 + x1) ≠ max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

• Proof step 32:

  max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(x2)), x2 + x1 ≠ [], max(x2 + x1) ≠ (max(x2) < a ⊃ a;max(x2)), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ x2 = [] ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used terms: x2, a

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

Lemma max-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  max(a ' + x0) = a, max(x1) ≠ a, x1 ≠ [], a < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ [] + x = x; (s) ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss) ⊦ x + [] = x; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

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

• Proof step 22:

  (max(x0) < a ⊃ a;max(x0)) = a, max(x1) ≠ a, x0 ≠ [], x1 ≠ [], a < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a; ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ x ⊑ [] ↔ x = []; (s) Used terms: x0, a

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

Lemma max-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  (max(x0) < a ⊃ a;max(x0)) = a, max(x1) ≠ a, x0 ≠ [], x1 ≠ [], a < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 24:

  a = a, max(x1) ≠ a, x0 ≠ [], x1 ≠ [], a < max(x1), max(x0) < a, ordered≤(x0), x1 ⊑ x0 ⊦

• Proof step 29:

  max(x0) = a, max(x1) ≠ a, x0 ≠ [], x1 ≠ [], a < max(x1), ordered≤(x0), x1 ⊑ x0, ¬ max(x0) < a ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  max(x1) ≠ a, x0 ≠ [], x1 ≠ [], max(x0) < a, a < max(x1), max(x0) < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ ordered≤(x) ↔ (∀ m, n. m ≤ n ∧ n < # x → ¬ x[n] < x[m])

• Proof step 27:

  max(x1) ≠ a, x0 ≠ [], x1 ≠ [], max(x0) < max(x1), a < max(x1), max(x0) < a, x1 ⊑ x0, ∀ m, n. m ≤ n ∧ n < # x0 → ¬ x0[n] < x0[m] ⊦

Rule argument

Additional information

Used terms: x0

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

A spec-lemma

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 28:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

Rule argument

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0, x1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  max(x1) ≠ max(x0), x0 ≠ [], x1 ≠ [], max(x0) < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a < a; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 31:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

Rule argument

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0, x1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(x2)), x2 + x1 ≠ [], max(x2 + x1) ≠ (max(x2) < a ⊃ a;max(x2)), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 34:

  max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(x2)), x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2) < a, ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

• Proof step 47:

  max((a ' + x2) + x0) = (max(x2) < a ⊃ a;max(x2)), x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2) < a, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  max((a ' + x2) + x0) = a, x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2) < a, a < max(x2 + x1), max(x2) < max(x2 + x1), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 36:

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

• Proof step 37:

  (max(x2 + x0) < a ⊃ a;max(x2 + x0)) = a, x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x0, a

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

Lemma max-rec

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  (max(x2 + x0) < a ⊃ a;max(x2 + x0)) = a, x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2) < a, a < max(x2 + x1), max(x2) < max(x2 + x1), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 39:

  a = a, x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, max(x2 + x0) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

• Proof step 44:

  max(x2 + x0) = a, x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x0) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2 + x0) < a, max(x2) < a, a < max(x2 + x1), max(x2) < max(x2 + x1), max(x2 + x0) < max(x2 + x1), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 41:

  x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2 + x0) < max(x2 + x1), max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, max(x2 + x0) < a, ordered≤(x0), x1 ⊑ x0, max(x2 + x0) ≠ max(x2) ∨ max(x2 + x1) = max(x2) ∨ ¬ ordered≤(x0) ∨ ¬ x1 ⊑ x0 ⊦

Rule argument

Additional information

Used terms: x0, x1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ a, x2 ≠ [], max(x2 + x0) < a, max(x2) < a, a < max(x2 + x1), max(x2) < max(x2 + x1), max(x2 + x0) < max(x2 + x1), ordered≤(x0), x1 ⊑ x0, max(x2 + x0) ≠ max(x2) ∨ max(x2 + x1) = max(x2) ⊦

Rule argument

Additional information

Proof step 42 from proof max-ordered (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 43:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

Rule argument

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0, x1

Proof step 43 from proof max-ordered (previous step) (back to tree)

A spec-lemma

Proof step 44 from proof max-ordered (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 45:

  x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2 + x0), x2 ≠ [], max(x2) < max(x2 + x0), max(x2 + x0) < max(x2 + x1), max(x2) < max(x2 + x1), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a < a; (s)

Proof step 45 from proof max-ordered (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 46:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

Rule argument

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0, x1

Proof step 46 from proof max-ordered (previous step) (back to tree)

A spec-lemma

Proof step 47 from proof max-ordered (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  max((a ' + x2) + x0) = max(x2), x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2) < a, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Proof step 48 from proof max-ordered (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 49:

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

• Proof step 50:

  (max(x2 + x0) < a ⊃ a;max(x2 + x0)) = max(x2), x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2) < a, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x2 + x0, a

Proof step 49 from proof max-ordered (previous step) (back to tree)

Lemma max-rec

Proof step 50 from proof max-ordered (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 51:

  (max(x2 + x0) < a ⊃ a;max(x2 + x0)) = max(x2), x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2) < a, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Proof step 51 from proof max-ordered (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 52:

  max(x2 + x0) = max(x2), x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2 + x0) < a, ¬ max(x2) < a, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (max(x + x0) < max(x) ↔ false); a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 52 from proof max-ordered (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

  max(x2 + x0) = max(x2), x2 + x0 ≠ [], x2 + x1 ≠ [], max(x2 + x1) ≠ max(x2), x2 ≠ [], ordered≤(x0), x1 ⊑ x0, ¬ max(x2) < a, ¬ max(x2 + x1) < a, ∀ x, x3. max(x2 + x) ≠ max(x2) ∨ max(x2 + x3) = max(x2) ∨ ¬ ordered≤(x) ∨ ¬ x3 ⊑ x ⊦

Rule argument

Additional information

Proof step 53 from proof max-ordered (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: x0, x1