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Proof step 2 from proof max-rem-notsame (previous step)

⊦ bag -- n = ∅, bag.max = (bag -- n).max ↔ ¬(n = bag.max ∧ count(n, bag) = 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ bag -- n = ∅, bag.max = (bag -- n).max ↔ n ≠ bag.max ∨ count(bag.max, bag) ≠ 1

Rule argument

Additional information