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Proof step 2 from proof count-del (previous step)

⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(b, bag) ≠ 0 ⊃ 1;0)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, b. count(a, bag0 -- b) = count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a, (bag0 ++ a0) -- b) = count(a, bag0 ++ a0) - (a = b ∧ count(b, bag0 ++ a0) ≠ 0 ⊃ 1;0)

Rule argument

indvar: bag

Additional information

Used simplifier rules: del-empty: ⊦ ∅ -- a = ∅;%bag generated by ∅, ++;;%count-empty: ⊦ count(a, ∅) = 0; from specification bag-basic