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Proof step 3 from proof put-in (previous step)

⊦ a0 × d0 ∈ ax[a, d] ↔ a0 × d0 = a × d, a0 × d0 ∈ ax

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ a0, d0, a, d. (a0 × d0 ∈ ax0[a, d] ↔ a0 × d0 = a × d) ∨ a0 × d0 ∈ ax0 ⊦ a0 × d0 ∈ (p + ax0)[a, d] ↔ a0 × d0 = a × d, a0 × d0 ∈ p + ax0

Rule argument

indvar: ax

Additional information

Used simplifier rules: put-empty: ⊦ ([])[a, d] = (a × d)';%assoclist freely generated by [], +;;% ⊦ p ∈ p0 ' ↔ p = p0;% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;