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Proof step 5 from proof put-in (previous step)

∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 ∈ (p ' + ax0)[a, d] ↔ a0 × d0 = a × d, a0 × d0 = p, a0 × d0 ∈ ax0

User applied rule structural induction

And obtained the following new goals

Proof step 6:
∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 ∈ ((a1 × d1)' + ax0)[a, d] ↔ a0 × d0 = a × d, a0 × d0 = a1 × d1, a0 × d0 ∈ ax0

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;