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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
max(x) < x.last, a0 < x.last, a < x.last, max(x) < a0, ordered≤(a0 ' + x), ∀ a1. a1 < x.last → a1 < max(x) ⊦ x = [], a < a0

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem