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Proof step 4 from proof in-assoclist-01 (previous step)

∀ a. a ∈ ax0 → a × ax0[a] ∈ ax0 ⊦ a × (p ' + ax0)[a] = p, a × (p ' + ax0)[a] ∈ ax0, ¬ a ∈ p ' ∧ ¬ a ∈ ax0

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: p

Additional information

Used simplifier rules: pair freely generated by ×;;