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Proof step 10 from proof ordered-app-max (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a < x.last, a0 < x.last, max(x) < x.last, ordered≤(a0 ' + x), ∀ a1. a1 < x.last → a1 < max(x) ⊦ x = [], max(x) < a0, a < max(x), a < a0

Rule argument

Additional information

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