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Proof step 2 from proof inkeys-nc (previous step)

⊦ a ∈ (b × d0)' + ax ↔ a ∈ ax, a = b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: inkeys-n: ⊦ a ≠ b → (a ∈ (b × d0)' + ax ↔ a ∈ ax);%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]