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Proof step 4 from proof inone (
previous step
)
a ∈ (b × d0)' ↔ (∃ d. a × d = b × d0) ⊦ a ∈ (b × d0)' ↔ a = b
User applied rule case distinction
And closed the goal!
Rule argument
pos: left 1
Additional information
Used simplifier rules: inj: ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0; from specification pair