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Proof step 4 from proof union-diff-union-same (previous step)

⊦ a ∈ (s0 ∪ s1) \ (s2 ∪ s1) ↔ a ∈ s0 \ (s2 ∪ s1)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%union-diff: ⊦ ¬ a ∈ s0 ∧ ¬ a ∈ s1 → (a ∈ s \ (s0 ∪ s1) ↔ a ∈ s);%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%union-diff: ⊦ ¬ a ∈ s0 ∧ ¬ a ∈ s1 → (a ∈ s \ (s0 ∪ s1) ↔ a ∈ s);%unin: ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0);%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic