Theorem sbequ1

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993) Revise df-sb . (Revised by BJ, 22-Dec-2020)

	Ref	Expression
Assertion	sbequ1	$$⊢x=t\to \left(\phi \to \left[t/x\right]\phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		equeucl	$$⊢x=t\to \left(y=t\to x=y\right)$$
2		ax12v	$$⊢x=y\to \left(\phi \to \forall x\left(x=y\to \phi \right)\right)$$
3	1 2	syl6	$$⊢x=t\to \left(y=t\to \left(\phi \to \forall x\left(x=y\to \phi \right)\right)\right)$$
4	3	com23	$$⊢x=t\to \left(\phi \to \left(y=t\to \forall x\left(x=y\to \phi \right)\right)\right)$$
5	4	alrimdv	$$⊢x=t\to \left(\phi \to \forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\right)$$
6		df-sb	$$⊢\left[t/x\right]\phi \leftrightarrow \forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$
7	5 6	imbitrrdi	$$⊢x=t\to \left(\phi \to \left[t/x\right]\phi \right)$$