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Proof step 3 from proof bag-rm-add-eq (previous step)

⊦ (∀ a0. count(a0, (bag -- a) ++ a) = count(a0, bag)) ↔ a ∈ bag

User applied rule case distinction

And obtained the following new goals

Proof step 4:
∀ a0. count(a0, (bag -- a) ++ a) = count(a0, bag) ⊦ a ∈ bag
Proof step 5:
a ∈ bag ⊦ ∀ a0. count(a0, (bag -- a) ++ a) = count(a0, bag)

Rule argument

pos: right 1

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