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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1))
Proof step 9:
∀ a. count(a, bag1) ≤ count(a, bag0), ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a, bag0) = count(a, bag1)

Rule argument

spec: inst: name: count-subset seq: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1)) subst: [bag1, bag0] → [bag1, bag0] flags: rewrite, current

Additional information

Used terms: bag1, bag0