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Proof step 4 from proof insfirst (previous step)

∀ 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 ', a ∈ ax1

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ a, d, ax0, d0. (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0 ∨ a ∈ ax1 ⊦ (((a0 × d1)' + ax1) + (a × d)' + ax0)[a, d0] = ((a0 × d1)' + ax1) + (a × d0)' + ax0, a ∈ (a0 × d1)', a ∈ ax1

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;