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Proof step 11 from proof max-ge-every (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a

Additional information

Used terms: a