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Proof step 1 from proof max-ins (conclusion)
⊦ a < s.max ∧ s ≠ ∅ → (s ++ a).max = s.max
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
Additional information
Used simplifier rules: max-ifins: ⊦ (s ++ a).max = (s = ∅ ⊃ a;(a < s.max ⊃ s.max;a));