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Proof step 2 from proof diff-union-01 (previous step)

⊦ a ∈ (s \ s0) ∪ s1 ↔ a ∈ s1, a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ a ∈ s ∨ a ∈ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;