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Proof step 7 from proof subset-elem (previous step)

∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a, bag0) = 0, a ∈ bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ bag ↔ count(a, bag) ≠ 0
Proof step 9:
∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a, bag0) = 0, count(a, bag1) ≠ 0

Rule argument

spec: inst: name: in seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0 subst: [bag, a] → [bag1, a] flags: rewrite, current

Additional information

Used terms: bag1, a