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Proof step 1 from proof max-add-one-different (conclusion)

⊦ m = (bag ++ m).max ↔ ¬(bag ≠ ∅ ∧ m < bag.max)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));% ⊦ ¬ n < n; (s)% ⊦ m > n ↔ n < m; (s)