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Proof step 16 from proof max-rem-notsame (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
count(bag.max, bag ++ m) = 0, m < bag.max ⊦ bag = ∅, count(bag.max, bag) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic