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Proof step 2 from proof union (previous step)

∀ a, bag1. count(a, bag ∪ bag1) = count(a, bag) + count(a, bag1) ⊦ count(a, (bag ++ a0) ∪ bag1) = count(a, bag ++ a0) + count(a, bag1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a, bag1. count(a, bag ∪ bag1) = count(a, bag) + count(a, bag1) ⊦ count(a, (bag ∪ bag1) ++ a0) = count(a, bag ++ a0) + count(a, bag1)

Rule argument

Additional information

Used simplifier rules: union-rec: ⊦ (bag0 ++ a) ∪ bag1 = (bag0 ∪ bag1) ++ a;