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Proof step 4 from proof union-subset-inj (previous step)

⊦ (∀ a. count(a, bag1) ≤ count(a, bag2)) ↔ bag1 ⊆ bag2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: count-subset seq: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1)) subst: [bag1, bag0] → [bag2, bag1] flags: <all paths>

Additional information

Used terms: bag2, bag1