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Proof step 4 from proof lookup (previous step)

∀ a, d, ax1. (ax0 + (a × d)' + ax1)[a] = d ∨ a ∈ ax0 ⊦ ((p ' + ax0) + (a × d)' + ax1)[a] = d, a ∈ ax0, a ∈ p '

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;