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Proof step 3 from proof del-notin (previous step)

count(a, bag) = 0 ⊦ bag -- a = bag

User applied rule structural induction

And obtained the following new goals

Proof step 4:
count(a, bag0 ++ a0) = 0, ∀ a. count(a, bag0) = 0 → bag0 -- a = bag0 ⊦ (bag0 ++ a0) -- a = bag0 ++ a0

Rule argument

indvar: bag

Additional information

Used simplifier rules: del-empty: ⊦ ∅ -- a = ∅;%bag generated by ∅, ++;;% ⊦ count(a, ∅) = 0; (s)