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Proof step 6 from proof inrec (previous step)

a × d = p, ∀ a0, d2, b0, d3. a0 × d2 ∈ ax0 ∨ a0 = b0 → (∃ d1. a0 × d1 ∈ ax0 ∨ a0 = b0 ∧ d1 = d3) ⊦ ∃ d1. a × d1 = b × d0 ∨ a × d1 = p ∨ a × d1 ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information