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Proof step 6 from proof differente-notin (previous step)

⊦ count(a, bag0) = 0, a ∈ bag1, a ∈ bag0 \ bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ bag ↔ count(a, bag) ≠ 0
Proof step 8:
count(a, bag0 \ bag1) = 0 ⊦ count(a, bag0) = 0, a ∈ bag1

Rule argument

spec: <toplevel> name: in seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0 subst: [bag, a] → [bag0 \ bag1, a] flags: <all paths>

Additional information

Used terms: bag0 \ bag1, a