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Proof step 5 from proof ex (previous step)

⊦ (∃ d. a × d ∈ ax) ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∃ d. a × d ∈ ax ⊦ ∃ d, ay, az. ax = ay + (a × d)' + az
Proof step 11:
∃ d, ay, az. ax = ay + (a × d)' + az ⊦ ∃ d. a × d ∈ ax

Rule argument

pos: right 1

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