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Proof step 9 from proof inapp (previous step)

⊦ (∃ d. a × d ∈ ax ∨ a × d ∈ ay) ↔ (∃ d. a × d ∈ ax) ∨ (∃ d. a × d ∈ ay)

User applied rule case distinction

And obtained the following new goals

Proof step 10:
∃ d. a × d ∈ ax ∨ a × d ∈ ay ⊦ ∃ d. a × d ∈ ax, ∃ d. a × d ∈ ay
Proof step 14:
(∃ d. a × d ∈ ax) ∨ (∃ d. a × d ∈ ay) ⊦ ∃ d. a × d ∈ ax ∨ a × d ∈ ay

Rule argument

pos: right 1

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