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Proof step 1 from proof int-inse (conclusion)

⊦ s ∩ (s0 ++ a) = ∅ → s ∩ s0 = ∅ ∧ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int-ins: ⊦ s ∩ (s0 ++ a) = ∅ ↔ s ∩ s0 = ∅ ∧ ¬ a ∈ s;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;