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Proof step 2 from proof int-del-rew (previous step)

⊦ s ∩ (s0 -- a) = (s ∩ s0) -- a, a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)