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Proof step 2 from proof inrec (previous step)

a × d ∈ ax ∨ a = b ⊦ ∃ d1. a × d1 = b × d0 ∨ a × d1 ∈ ax

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, d, b, d0. a × d ∈ ax0 ∨ a = b → (∃ d1. a × d1 = b × d0 ∨ a × d1 ∈ ax0) ⊦ a × d ∈ p + ax0 ∨ a = b → (∃ d1. a × d1 = b × d0 ∨ a × d1 ∈ p + ax0)

Rule argument

term: ax

Additional information

Used simplifier rules: assoclist freely generated by [], +;;% ⊦ ¬ p ∈ [];% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;