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Proof step 3 from proof bag-rem-notmax (previous step)

⊦ (bag -- m).max = bag.max, m = bag.max

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ ((bag0 ++ m0) -- m).max = (bag0 ++ m0).max, m = (bag0 ++ m0).max

Rule argument

term: bag

Additional information

Used simplifier rules: nat-bag generated by ∅, ++;;% ⊦ ∅ -- n = ∅; (s)