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Proof step 3 from proof insfirst (previous step)

⊦ (ax + (a × d)' + ax0)[a, d0] = ax + (a × d0)' + ax0, a ∈ ax

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ a, d, ax0, d0. (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0 ∨ a ∈ ax1 ⊦ ((p + ax1) + (a × d)' + ax0)[a, d0] = (p + ax1) + (a × d0)' + ax0, a ∈ p + ax1

Rule argument

indvar: ax

Additional information

Used simplifier rules: put-yes: ⊦ ((a × d0)' + ax)[a, d] = (a × d)' + ax;%assoclist freely generated by [], +;;% ⊦ [] + ax = ax; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1; (s)