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

⊦ (ax + (a × d)')[a] = d, a ∈ ax

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, d. (ax0 + (a × d)')[a] = d ∨ a ∈ ax0 ⊦ ((p + ax0) + (a × d)')[a] = d ∨ a ∈ p + ax0

Rule argument

term: ax

Additional information

Used simplifier rules: one-01: ⊦ (a × d)'[a] = d;%assoclist freely generated by [], +;;% ⊦ [] + ax = ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;