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Proof step 3 from proof in-put (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ a, b, d. a ∈ ax0[b, d] ↔ a = b ∨ a ∈ ax0 ⊦ a ∈ ((a0 × d0)' + ax0)[b, d] ↔ a = b ∨ a ∈ (a0 × d0)' ∨ a ∈ ax0

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;