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Proof step 8 from proof bag-rm-add-eq (previous step)

⊦ (a0 = a ⊃ 1;0) + count(a0, bag -- a) = count(a0, bag), count(a, bag) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ count(a, bag) = 0, (a0 = a ⊃ 1;0) + (count(a0, bag) - (a0 = a ⊃ 1;0)) = count(a0, bag)

Rule argument

Additional information

Used simplifier rules: count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);