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Proof step 2 from proof in-03 (previous step)

a ∈ ax + ay ⊦ a ∈ ax + p ' + ay

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay;%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]