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Proof step 7 from proof max-ordered (previous step)

max((a ' + x2) + x0) = max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x0, x1. max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → x2 = [] ∨ max(x2 + x1) = max(x2) ⊦ (max(x2 + x1) < a ⊃ a;max(x2 + x1)) = max(a ' + x2), x2 + x1 = []

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
max(x2 + x1) < a, max((a ' + x2) + x0) = max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x0, x1. max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → x2 = [] ∨ max(x2 + x1) = max(x2) ⊦ a = max(a ' + x2), x2 + x1 = []
Proof step 15:
¬ max(x2 + x1) < a, max((a ' + x2) + x0) = max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x0, x1. max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → x2 = [] ∨ max(x2 + x1) = max(x2) ⊦ max(x2 + x1) = max(a ' + x2), x2 + x1 = []

Rule argument

pos: right 1

Proof step 7 from proof max-ordered (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information