Theorem sbequ8

Description: Elimination of equality from antecedent after substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018) Revise df-sb . (Revised by Wolf Lammen, 28-Jul-2023) (New usage is discouraged.)

	Ref	Expression
Assertion	sbequ8	$$⊢\left[y/x\right]\phi \leftrightarrow \left[y/x\right]\left(x=y\to \phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		equsb1	$$⊢\left[y/x\right]x=y$$
2	1	a1bi	$$⊢\left[y/x\right]\phi \leftrightarrow \left(\left[y/x\right]x=y\to \left[y/x\right]\phi \right)$$
3		sbim	$$⊢\left[y/x\right]\left(x=y\to \phi \right)\leftrightarrow \left(\left[y/x\right]x=y\to \left[y/x\right]\phi \right)$$
4	2 3	bitr4i	$$⊢\left[y/x\right]\phi \leftrightarrow \left[y/x\right]\left(x=y\to \phi \right)$$