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Proof step 11 from proof inrec (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: d0

Additional information

Used terms: d0