Theorem bj-ssblem1

Description: A lemma for the definiens of df-sb . An instance of sp proved without it. Note: it has a common subproof with sbjust . (Contributed by BJ, 22-Dec-2020) (Proof modification is discouraged.)

	Ref	Expression
Assertion	bj-ssblem1	$$⊢\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\to \left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		equequ1	$$⊢y=z\to \left(y=t\leftrightarrow z=t\right)$$
2		equequ2	$$⊢y=z\to \left(x=y\leftrightarrow x=z\right)$$
3	2	imbi1d	$$⊢y=z\to \left(\left(x=y\to \phi \right)\leftrightarrow \left(x=z\to \phi \right)\right)$$
4	3	albidv	$$⊢y=z\to \left(\forall x\left(x=y\to \phi \right)\leftrightarrow \forall x\left(x=z\to \phi \right)\right)$$
5	1 4	imbi12d	$$⊢y=z\to \left(\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\leftrightarrow \left(z=t\to \forall x\left(x=z\to \phi \right)\right)\right)$$
6	5	spw	$$⊢\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\to \left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$