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	$$⊢CondEq\left(x=y\to \left(\phi \leftrightarrow \psi \right)\right)$$
Assertion	cdeqab1	$$⊢CondEq\left(x=y\to \left\{x|\phi \right\}=\left\{y|\psi \right\}\right)$$

Proof

Step	Hyp	Ref	Expression
1		cdeqnot.1	$$⊢CondEq\left(x=y\to \left(\phi \leftrightarrow \psi \right)\right)$$
2		nfv	$$⊢Ⅎy\phi$$
3		nfv	$$⊢Ⅎx\psi$$
4	1	cdeqri	$$⊢x=y\to \left(\phi \leftrightarrow \psi \right)$$
5	2 3 4	cbvab	$$⊢\left\{x|\phi \right\}=\left\{y|\psi \right\}$$
6	5	cdeqth	$$⊢CondEq\left(x=y\to \left\{x|\phi \right\}=\left\{y|\psi \right\}\right)$$