Theorem bj-sbel1

Description: Version of sbcel1g when substituting a set. (Note: one could have a corresponding version of sbcel12 when substituting a set, but the point here is that the antecedent of sbcel1g is not needed when substituting a set.) (Contributed by BJ, 6-Oct-2018)

	Ref	Expression
Assertion	bj-sbel1	$$⊢\left[y/x\right]A\in B\leftrightarrow ⦋y/x⦌A\in B$$

Proof

Step	Hyp	Ref	Expression
1		sbsbc	$$⊢\left[y/x\right]A\in B\leftrightarrow \underset{\dot{}}{[}y/x\underset{\dot{}}{]}A\in B$$
2		sbcel1g	$$⊢y\in \mathrm{V}\to \left(\underset{\dot{}}{[}y/x\underset{\dot{}}{]}A\in B\leftrightarrow ⦋y/x⦌A\in B\right)$$
3	2	elv	$$⊢\underset{\dot{}}{[}y/x\underset{\dot{}}{]}A\in B\leftrightarrow ⦋y/x⦌A\in B$$
4	1 3	bitri	$$⊢\left[y/x\right]A\in B\leftrightarrow ⦋y/x⦌A\in B$$