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Proof step 5 from proof difference (previous step)

⊦ count(a, bag0 \ ∅) = (count(a, ∅) < count(a, bag0) ⊃ count(a, bag0) - count(a, ∅);0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ count(a, bag0) = (count(a, bag0) ≠ 0 ⊃ count(a, bag0);0)

Rule argument

Additional information

Used simplifier rules: diff-empty: ⊦ bag0 \ ∅ = bag0;%sub-base: ⊦ m - 0 = m; from specification nat%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%count-empty: ⊦ count(a, ∅) = 0; from specification bag-basic