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Proof step 1 from proof in-pair (conclusion)
⊦ a × d ∈ st ↔ ¬ ¬(a ∈ st ∧ st[a] = d)
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
Additional information
Used simplifier rules: In-store: ⊦ a × d ∈ st ↔ a ∈ st ∧ st[a] = d;