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Proof step 3 from proof union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ≠ b → count(a, bag ++ b) = count(a, bag)
Proof step 5:
¬ a ≠ a0, ∀ a, bag1. count(a, bag ∪ bag1) = count(a, bag) + count(a, bag1) ⊦ count(a, (bag ∪ bag1) ++ a0) = count(a, bag ++ a0) + count(a, bag1)
Proof step 7:
a ≠ a0, ∀ a, bag1. count(a, bag ∪ bag1) = count(a, bag) + count(a, bag1) ⊦ count(a, bag ∪ bag1) = count(a, bag ++ a0) + count(a, bag1)

Rule argument

spec: bag-basic inst: name: count-other seq: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag) subst: [b, bag, a] → [a0, bag ∪ bag1, a] flags: rewrite

Additional information

Used terms: a0, bag ∪ bag1, a