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Proof step 8 from proof max-ge-every (previous step)

(max(x0) < a0 ⊃ a0;max(x0)) < a, a ∈ x0, ∀ a1. ¬ max(x0) < a1 ∨ ¬ a1 ∈ x0 ⊦ a = a0, x0 = []

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
a0 < a, max(x0) < a0, a ∈ x0, ∀ a1. ¬ max(x0) < a1 ∨ ¬ a1 ∈ x0 ⊦ a = a0, x0 = []
Proof step 11:
max(x0) < a, a ∈ x0, ∀ a1. ¬ max(x0) < a1 ∨ ¬ a1 ∈ x0 ⊦ a = a0, x0 = [], max(x0) < a0

Rule argument

pos: left 1

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