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Proof step 2 from proof union-one (previous step)

⊦ a ∈ {b} ∪ s ↔ a ∈ s, a = b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: One: ⊦ {a} = ∅ ++ a;%unadd: ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a;%une: ⊦ s ∪ ∅ = s;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws)