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Proof step 3 from proof in-01 (previous step)

a ∈ ax ↔ (∃ d. a × d ∈ ax) ⊦ a × d ∈ ax → a ∈ ax

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a × d ∈ ax ⊦ a ∈ ax, ∃ d. a × d ∈ ax

Rule argument

Additional information