Theorem sblbis

Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993)

	Ref	Expression
Hypothesis	sblbis.1	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
Assertion	sblbis	$$⊢\left[y/x\right]\left(\chi \leftrightarrow \phi \right)\leftrightarrow \left(\left[y/x\right]\chi \leftrightarrow \psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sblbis.1	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
2		sbbi	$$⊢\left[y/x\right]\left(\chi \leftrightarrow \phi \right)\leftrightarrow \left(\left[y/x\right]\chi \leftrightarrow \left[y/x\right]\phi \right)$$
3	1	bibi2i	$$⊢\left(\left[y/x\right]\chi \leftrightarrow \left[y/x\right]\phi \right)\leftrightarrow \left(\left[y/x\right]\chi \leftrightarrow \psi \right)$$
4	2 3	bitri	$$⊢\left[y/x\right]\left(\chi \leftrightarrow \phi \right)\leftrightarrow \left(\left[y/x\right]\chi \leftrightarrow \psi \right)$$