Theorem sbrbis

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993)

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

Proof

Step	Hyp	Ref	Expression
1		sbrbis.1	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
2		sbbi	$$⊢\left[y/x\right]\left(\phi \leftrightarrow \chi \right)\leftrightarrow \left(\left[y/x\right]\phi \leftrightarrow \left[y/x\right]\chi \right)$$
3	1	bibi1i	$$⊢\left(\left[y/x\right]\phi \leftrightarrow \left[y/x\right]\chi \right)\leftrightarrow \left(\psi \leftrightarrow \left[y/x\right]\chi \right)$$
4	2 3	bitri	$$⊢\left[y/x\right]\left(\phi \leftrightarrow \chi \right)\leftrightarrow \left(\psi \leftrightarrow \left[y/x\right]\chi \right)$$