Theorem sbequiOLD

Description: Obsolete version of sbequ as of 10-Aug-2026, as per conventions , paragraph "Biconditional". (Contributed by NM, 14-May-1993) (Proof shortened by Wolf Lammen, 15-Sep-2018) (Proof shortened by Steven Nguyen, 7-Jul-2023) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	sbequiOLD	$$⊢x=y\to \left(\left[x/z\right]\phi \to \left[y/z\right]\phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sbequ	$$⊢x=y\to \left(\left[x/z\right]\phi \leftrightarrow \left[y/z\right]\phi \right)$$
2	1	biimpd	$$⊢x=y\to \left(\left[x/z\right]\phi \to \left[y/z\right]\phi \right)$$