Theorem sbequOLD

Description: Obsolete proof of sbequ as of 7-Jul-2023. An equality theorem for substitution. Used in proof of Theorem 9.7 in Megill p. 449 (p. 16 of the preprint). (Contributed by NM, 14-May-1993) (Proof modification is discouraged.) (New usage is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		sbequi	$$⊢x=y\to \left(\left[x/z\right]\phi \to \left[y/z\right]\phi \right)$$
2		sbequi	$$⊢y=x\to \left(\left[y/z\right]\phi \to \left[x/z\right]\phi \right)$$
3	2	equcoms	$$⊢x=y\to \left(\left[y/z\right]\phi \to \left[x/z\right]\phi \right)$$
4	1 3	impbid	$$⊢x=y\to \left(\left[x/z\right]\phi \leftrightarrow \left[y/z\right]\phi \right)$$