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Proof step 2 from proof insnotin (previous step)

⊦ ax[a, d] = ax + (a × d)', a ∈ ax

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: ax

Additional information

Used simplifier rules: put-empty: ⊦ ([])[a, d] = (a × d)';%assoclist freely generated by [], +;;%Nil: ⊦ [] + ax = ax; from specification list[assoclista]%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]%c-04: ⊦ p ' = p0 ' ↔ p = p0; from specification list[assoclista]