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Proof step 2 from proof exin (previous step)

⊦ count(a, bag) = 0, ∃ bag0. bag = bag0 ++ a

User applied rule exists right

And obtained the following new goals

Proof step 3:
⊦ count(a, bag) = 0, bag = (bag -- a) ++ a, ∃ bag0. bag = bag0 ++ a

Rule argument

pos: right 2 terms: bag -- a abort: F discard: F computed: F

Additional information

Used terms: bag -- a