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Proof step 1 from proof ls-max (conclusion)

⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(n < bag.max ⊃ bag.max;n))

Rule argument

Additional information

Used simplifier rules: ⊦ m > n ↔ n < m; (s)