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Proof step 1 from proof in-rec (conclusion)
⊦ n ∈ bag ++ m ↔ n ∈ bag ∨ n = m
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 2
:
⊦ n ∈ bag0 ↔ count(n, bag0) ≠ 0
Proof step 3
:
⊦ count(n, bag ++ m) ≠ 0 ↔ n ∈ bag ∨ n = m
Rule argument
spec: bag-union, inst: nat-bag name: in seq: ⊦ n ∈ bag0 ↔ count(n, bag0) ≠ 0 subst: [bag0, n] → [bag ++ m, n] flags: <all paths>
Additional information
Used terms: bag ++ m, n