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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. ¬ a ∈ ax ∧ p ' + ax0 = ax + (a × d)' + ax1

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)) → ¬ a0 ∈ (a × d)' ∧ ¬ a0 ∈ ax0 ∨ (∃ ax, d0, ax1. ¬ a0 ∈ ax ∧ (a × d)' + ax0 = ax + (a0 × d0)' + ax1)

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;