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Proof step 3 from proof Diff-Diff-com (previous step)

⊦ ∀ a. (a ∈ (st \ s) \ s0 ↔ a ∈ (st \ s0) \ s) ∧ (a ∈ (st \ s) \ s0 → ((st \ s) \ s0)[a] = ((st \ s0) \ s)[a])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information