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Proof step 2 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 3:
∀ a, d, ax1, d0. (ax0 + (a × d)' + ax1)[a, d0] = ax0 + (a × d0)' + ax1 ∨ a ∈ ax0 ⊦ ((p + ax0) + (a × d)' + ax1)[a, d0] = (p + ax0) + (a × d0)' + ax1 ∨ a ∈ p + ax0

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;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1;