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Proof step 7 from proof bag-swap-del (previous step)

⊦ a0 = b, count(a0, bag -- a -- b) = count(a0, bag -- b -- a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ a0 = b, count(a0, bag -- a) = count(a0, bag -- b -- a)

Rule argument

Additional information

Used simplifier rules: count-del-other: ⊦ a ≠ b → count(a, bag -- b) = count(a, bag);