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Proof step 4 from proof bag-rm-add-eq (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 5:
count(a, (bag -- a) ++ a) = count(a, bag) ⊦ a ∈ bag

Rule argument

pos: left 1 terms: a abort: F discard: T computed: T

Additional information

Used terms: a