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Proof step 4 from proof subset-elem (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ bag ↔ count(a, bag) ≠ 0
Proof step 6:
∀ a. count(a, bag0) ≤ count(a, bag1), count(a, bag0) ≠ 0 ⊦ a ∈ bag1

Rule argument

spec: inst: name: in seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0 subst: [bag, a] → [bag0, a] flags: rewrite, current

Additional information

Used terms: bag0, a