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Proof step 6 from proof del-notin (previous step)

count(a, bag0 ++ a0) = 0, ∀ a. count(a, bag0) = 0 → bag0 -- a = bag0 ⊦ (bag0 -- a) ++ a0 = bag0 ++ a0, a0 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
count(a, bag0) = 0, ∀ a. count(a, bag0) = 0 → bag0 -- a = bag0 ⊦ (bag0 -- a) ++ a0 = bag0 ++ a0, a0 = a

Rule argument

Additional information

Used simplifier rules: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); (s)