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Proof step 5 from proof still-in-rem (previous step)

⊦ count(a, bag0 -- b) = 0, a ∈ bag0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) = 0, a ∈ bag0

Rule argument

Additional information

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