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Proof step 2 from proof dom-max-in (previous step)

n > (dom st).max ⊦ n ∈ st ↔ false, st = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
(dom st).max < n, n ∈ st ⊦ st = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m > n ↔ n < m; (s)