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Proof step 6 from proof max-append (previous step)

x1 + x0 ≠ [], ∀ x0. max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ∨ x1 + x0 = [] ⊦ (max(x1 + x0) < a ⊃ a;max(x1 + x0)) = max(a ' + x1), (max(x1 + x0) < a ⊃ a;max(x1 + x0)) = max(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ x0. max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ∨ x1 + x0 = [] ⊦ (max(x1 + x0) < a ⊃ a;max(x1 + x0)) = max(a ' + x1), (max(x1 + x0) < a ⊃ a;max(x1 + x0)) = max(x0), x1 + x0 = []

Proof step 6 from proof max-append (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information