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Proof step 3 from proof inone (previous step)

a ∈ (b × d0)' ↔ (∃ d. a × d ∈ (b × d0)') ⊦ a ∈ (b × d0)' ↔ a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ (b × d0)' ↔ (∃ d. a × d = b × d0) ⊦ a ∈ (b × d0)' ↔ a = b

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ p ∈ p0 ' ↔ p = p0; from specification list[assoclista]