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Proof step 4 from proof del-notin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
count(a, bag0 ++ a0) = 0, ∀ a. count(a, bag0) = 0 → bag0 -- a = bag0 ⊦ (a0 = a ⊃ bag0;(bag0 -- a) ++ a0) = bag0 ++ a0

Rule argument

Additional information

Used simplifier rules: del-rec: ⊦ (bag ++ a) -- b = (a = b ⊃ bag;(bag -- b) ++ a);