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Proof step 1 from proof subset-antisym (conclusion)

⊦ bag0 ⊆ bag1 → (bag1 ⊆ bag0 ↔ bag0 = bag1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bag1 ⊆ bag0, bag0 ⊆ bag1 ⊦ bag0 = bag1

Rule argument

Additional information

Used simplifier rules: subset-trans: ⊦ bag0 ⊆ bag1 ∧ bag1 ⊆ bag2 → bag0 ⊆ bag2;%subset-refl: ⊦ bag ⊆ bag;