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Proof step 1 from proof undel-03 (conclusion)

⊦ ¬ a ∈ s → s ∪ (s0 -- a) = (s ∪ s0) -- a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: undel-01: ⊦ (s ∪ s0) -- a = s ∪ (s0 -- a) ↔ ¬ a ∈ s;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;