Theorem cdeqab

Description: Distribute conditional equality over abstraction. (Contributed by Mario Carneiro, 11-Aug-2016)

	Ref	Expression
Hypothesis	cdeqnot.1	⊢ CondEq ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion	cdeqab	⊢ CondEq ( 𝑥 = 𝑦 → { 𝑧 ∣ 𝜑 } = { 𝑧 ∣ 𝜓 } )

Proof

Step	Hyp	Ref	Expression
1		cdeqnot.1	⊢ CondEq ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2	1	cdeqri	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
3	2	abbidv	⊢ ( 𝑥 = 𝑦 → { 𝑧 ∣ 𝜑 } = { 𝑧 ∣ 𝜓 } )
4	3	cdeqi	⊢ CondEq ( 𝑥 = 𝑦 → { 𝑧 ∣ 𝜑 } = { 𝑧 ∣ 𝜓 } )