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Proof step 2 from proof last-01 (previous step)

⊦ disj(y + b ', x) ↔ disj(x, y), b ∈ x

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: app-03: ⊦ disj(y + z, x) ↔ disj(x, y) ∧ disj(x, z);%one-01: ⊦ disj(x, a ') ↔ ¬ a ∈ x;%a: ⊦ (x + y) + z = x + y + z; from specification list