Theorem sbrbif

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993) (Revised by Mario Carneiro, 4-Oct-2016)

	Ref	Expression
Hypotheses	sbrbif.1	$$⊢Ⅎx\chi$$
	sbrbif.2	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
Assertion	sbrbif	$$⊢\left[y/x\right]\left(\phi \leftrightarrow \chi \right)\leftrightarrow \left(\psi \leftrightarrow \chi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sbrbif.1	$$⊢Ⅎx\chi$$
2		sbrbif.2	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
3	2	sbrbis	$$⊢\left[y/x\right]\left(\phi \leftrightarrow \chi \right)\leftrightarrow \left(\psi \leftrightarrow \left[y/x\right]\chi \right)$$
4	1	sbf	$$⊢\left[y/x\right]\chi \leftrightarrow \chi$$
5	4	bibi2i	$$⊢\left(\psi \leftrightarrow \left[y/x\right]\chi \right)\leftrightarrow \left(\psi \leftrightarrow \chi \right)$$
6	3 5	bitri	$$⊢\left[y/x\right]\left(\phi \leftrightarrow \chi \right)\leftrightarrow \left(\psi \leftrightarrow \chi \right)$$