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Proof step 14 from proof max-rem-notsame (previous step)

count((bag ++ m).max, bag ++ m) = 0, count(bag.max, bag) = 0 → bag = ∅ ⊦ bag ++ m = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
count((m < bag.max ⊃ bag.max;m), bag ++ m) = 0 ⊦ bag = ∅, count(bag.max, bag) = 0

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));% ⊦ count(n, ∅ ++ n) = 1; (s)% ⊦ bag0 ++ n ≠ ∅; (s)% ⊦ m > n ↔ n < m; (s)