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

⊦ count(a, ∅ ++ b) = (a = b ⊃ 1;0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ≠ b → count(a, bag ++ b) = count(a, bag)

Rule argument

spec: bag-basic inst: name: count-other seq: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag) subst: [b, bag, a] → [b, ∅, a] flags: rewrite

Additional information

Used simplifier rules: count-empty: ⊦ count(a, ∅) = 0; from specification bag-basic%count-1: ⊦ count(a, ∅ ++ a) = 1; from specification bag-basic Used terms: b, ∅, a