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Proof step 1 from proof subset-elem (conclusion)

⊦ bag0 ⊆ bag1 ∧ a ∈ bag0 → (¬ a ∈ bag1 ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bag0 ⊆ bag1, a ∈ bag0 ⊦ a ∈ bag1

Rule argument

Additional information