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Proof step 4 from proof insert-nonempty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a0. (a0 = a ⊃ 1;0) + count(a0, bag) = 0 ⊦

Rule argument

Additional information

Used simplifier rules: count-insert: ⊦ count(a, bag ++ b) = (a = b ⊃ 1;0) + count(a, bag);%count-empty: ⊦ count(a, ∅) = 0;