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Proof step 4 from proof firstoc (previous step)

∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2), a ∈ p ' ∨ a ∈ ax0 ⊦ ∃ ax, d, ax1. p ' + ax0 = ax + (a × d)' + ax1 ∧ ¬ a ∈ ax

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2), a ∈ (a0 × d)' ∨ a ∈ ax0 ⊦ ∃ ax, d0, ax1. (a0 × d)' + ax0 = ax + (a × d0)' + ax1 ∧ ¬ a ∈ ax

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;