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Proof step 11 from proof subset-one (previous step)

count(a0, bag) < count(a0, ∅ ++ a) ⊦ count(a, bag) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ 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] → [a, ∅, a0] flags: rewrite

Additional information

Used simplifier rules: count-empty: ⊦ count(a, ∅) = 0; from specification bag-basic%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%count-1: ⊦ count(a, ∅ ++ a) = 1; from specification bag-basic%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1 Used terms: a, ∅, a0