Theorem cdeqab1

Description: Distribute conditional equality over abstraction. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by Mario Carneiro, 11-Aug-2016) (New usage is discouraged.)

	Ref	Expression
Hypothesis	cdeqnot.1	⊢ CondEq ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion	cdeqab1	⊢ CondEq ( 𝑥 = 𝑦 → { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 } )

Proof

Step	Hyp	Ref	Expression
1		cdeqnot.1	⊢ CondEq ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2		nfv	⊢ Ⅎ 𝑦 𝜑
3		nfv	⊢ Ⅎ 𝑥 𝜓
4	1	cdeqri	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
5	2 3 4	cbvab	⊢ { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 }
6	5	cdeqth	⊢ CondEq ( 𝑥 = 𝑦 → { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 } )