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

(max(x1 + x0) < a ⊃ a;max(x1 + x0)) < max(a ' + x1), (∃ x0. max(x1 + x0) < max(x1)) → x1 = [] ⊦ x1 + x0 = []

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
max(x1 + x0) < a, a < max(a ' + x1), (∃ x0. max(x1 + x0) < max(x1)) → x1 = [] ⊦ x1 + x0 = []
Proof step 13:
¬ max(x1 + x0) < a, max(x1 + x0) < max(a ' + x1), (∃ x0. max(x1 + x0) < max(x1)) → x1 = [] ⊦ x1 + x0 = []

Rule argument

pos: left 1

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