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Proof step 4 from proof some (previous step)

⊦ ∀ a. count(a, bag) = count(a, ∅), ∃ a. count(a, bag) ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ count(a, bag) = 0, ∃ a. count(a, bag) ≠ 0

Rule argument

Additional information

Used simplifier rules: count-empty: ⊦ count(a, ∅) = 0; from specification bag-basic