Theorem eqsnd

Description: Deduce that a set is a singleton. (Contributed by Thierry Arnoux, 10-May-2023) (Proof shortened by SN, 3-Jul-2025)

	Ref	Expression
Hypotheses	eqsnd.1	|- ( ( ph /\ x e. A ) -> x = B )
	eqsnd.2	|- ( ph -> B e. A )
Assertion	eqsnd	|- ( ph -> A = { B } )

Proof

Step	Hyp	Ref	Expression
1		eqsnd.1	|- ( ( ph /\ x e. A ) -> x = B )
2		eqsnd.2	|- ( ph -> B e. A )
3	1	ralrimiva	|- ( ph -> A. x e. A x = B )
4	2	ne0d	|- ( ph -> A =/= (/) )
5		eqsn	|- ( A =/= (/) -> ( A = { B } <-> A. x e. A x = B ) )
6	4 5	syl	|- ( ph -> ( A = { B } <-> A. x e. A x = B ) )
7	3 6	mpbird	|- ( ph -> A = { B } )