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Proof step 22 from proof max-ordered (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 23:
max(x0) < a, a = a, ordered≤(x0), x1 ⊑ x0, a < max(x1) ⊦ x0 = [], max(x1) = a, x1 = []
Proof step 28:
¬ max(x0) < a, max(x0) = a, ordered≤(x0), x1 ⊑ x0, a < max(x1) ⊦ x0 = [], max(x1) = a, x1 = []

Rule argument

pos: left 1

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