Theorem ceqsex2v

Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006) Avoid ax-10 and ax-11 . (Revised by GG, 20-Aug-2023)

	Ref	Expression
Hypotheses	ceqsex2v.1	|- A e. _V
	ceqsex2v.2	|- B e. _V
	ceqsex2v.3	|- ( x = A -> ( ph <-> ps ) )
	ceqsex2v.4	|- ( y = B -> ( ps <-> ch ) )
Assertion	ceqsex2v	|- ( E. x E. y ( x = A /\ y = B /\ ph ) <-> ch )

Proof

Step	Hyp	Ref	Expression
1		ceqsex2v.1	|- A e. _V
2		ceqsex2v.2	|- B e. _V
3		ceqsex2v.3	|- ( x = A -> ( ph <-> ps ) )
4		ceqsex2v.4	|- ( y = B -> ( ps <-> ch ) )
5		3anass	|- ( ( x = A /\ y = B /\ ph ) <-> ( x = A /\ ( y = B /\ ph ) ) )
6	5	exbii	|- ( E. y ( x = A /\ y = B /\ ph ) <-> E. y ( x = A /\ ( y = B /\ ph ) ) )
7		19.42v	|- ( E. y ( x = A /\ ( y = B /\ ph ) ) <-> ( x = A /\ E. y ( y = B /\ ph ) ) )
8	6 7	bitri	|- ( E. y ( x = A /\ y = B /\ ph ) <-> ( x = A /\ E. y ( y = B /\ ph ) ) )
9	8	exbii	|- ( E. x E. y ( x = A /\ y = B /\ ph ) <-> E. x ( x = A /\ E. y ( y = B /\ ph ) ) )
10	3	anbi2d	|- ( x = A -> ( ( y = B /\ ph ) <-> ( y = B /\ ps ) ) )
11	10	exbidv	|- ( x = A -> ( E. y ( y = B /\ ph ) <-> E. y ( y = B /\ ps ) ) )
12	1 11	ceqsexv	|- ( E. x ( x = A /\ E. y ( y = B /\ ph ) ) <-> E. y ( y = B /\ ps ) )
13	2 4	ceqsexv	|- ( E. y ( y = B /\ ps ) <-> ch )
14	9 12 13	3bitri	|- ( E. x E. y ( x = A /\ y = B /\ ph ) <-> ch )