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Proof step 5 from proof subset-elem (previous step)

a ∈ bag0, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ a ∈ bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ bag ↔ count(a, bag) ≠ 0
Proof step 7:
∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a, bag0) = 0, a ∈ bag1

Rule argument

spec: <toplevel> name: in seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0 subst: [bag, a] → [bag0, a] flags: <all paths>

Additional information

Used terms: bag0, a