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Proof step 8 from proof ordered-app-max (previous step)

max(x) < x.last, a0 < x.last, a < x.last, ordered≤(a0 ' + x), max(x) < a0, ∀ a1. a1 < x.last → a1 < max(x) ⊦ x = [], a < a0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: max(x)

Additional information

Used simplifier rules: irreflexivity: ⊦ ¬ a < a; from specification oelem Used terms: max(x)