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Proof step 5 from proof bag-rm-add-eq (previous step)

a ∈ bag ⊦ ∀ a0. count(a0, (bag -- a) ++ a) = count(a0, bag)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ count(a0, (bag -- a) ++ a) = count(a0, bag), count(a, bag) = 0

Rule argument

Additional information

Used simplifier rules: in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;