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Proof step 5 from proof max-rem-notsame (previous step)

bag.max = (bag -- bag.max).max, count(bag.max, bag) = 1 ⊦ bag -- bag.max = ∅

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 6:
(bag0 ++ m).max = ((bag0 ++ m) -- (bag0 ++ m).max).max, count((bag0 ++ m).max, bag0 ++ m) = 1, bag0.max = (bag0 -- bag0.max).max ∧ count(bag0.max, bag0) = 1 → bag0 -- bag0.max = ∅ ⊦ (bag0 ++ m) -- (bag0 ++ m).max = ∅

Rule argument

term: bag

Additional information

Used simplifier rules: nat-bag generated by ∅, ++;;% ⊦ count(n, ∅) = 0; (s)