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Proof step 1 from proof diff-union-02 (conclusion)

⊦ ¬ a ∈ s → (a ∈ ((s ∪ s2) \ s0) ∪ s1 ↔ a ∈ (s2 \ s0) ∪ s1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a ∈ (s ∪ s2) \ s0 ↔ a ∈ s2 \ s0, a ∈ s, a ∈ s1

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;