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Proof step 2 from proof firstoc (previous step)

a ∈ ax ⊦ ∃ ax0, d, ax1. ax = ax0 + (a × d)' + ax1 ∧ ¬ a ∈ ax0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2), a ∈ p + ax0 ⊦ ∃ ax2, d, ax1. p + ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2

Rule argument

indvar: ax

Additional information

Used simplifier rules: empty: ⊦ ¬ a ∈ [];%assoclist freely generated by [], +;;