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Proof step 11 from proof inapp (previous step)

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 12:
a × d ∈ ax ⊦ ∃ d. a × d ∈ ax, ∃ d. a × d ∈ ay
Proof step 13:
a × d ∈ ay ⊦ ∃ d. a × d ∈ ax, ∃ d. a × d ∈ ay

Rule argument

pos: left 1

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