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Proof step 10 from proof max-ge-every (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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