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Proof step 1 from proof exin (conclusion)

⊦ a ∈ bag → (∃ bag0. bag = bag0 ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ count(a, bag) = 0, ∃ bag0. bag = bag0 ++ a

Rule argument

Additional information

Used simplifier rules: in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;