Theorem sblbisvOLD

Description: Obsolete version of sblbis as of 24-Jul-2023. Introduce left biconditional inside of a substitution. Version of sblbis with a disjoint variable condition, not requiring ax-13 . (Contributed by Wolf Lammen, 18-Jan-2023) (New usage is discouraged.) (Proof modification is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		sblbisvOLD.1	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
2		sbbivOLD	$$⊢\left[y/x\right]\left(\chi \leftrightarrow \phi \right)\leftrightarrow \left(\left[y/x\right]\chi \leftrightarrow \left[y/x\right]\phi \right)$$
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	$$⊢\left[y/x\right]\left(\chi \leftrightarrow \phi \right)\leftrightarrow \left(\left[y/x\right]\chi \leftrightarrow \psi \right)$$