Theorem ceqsex2

Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006)

	Ref	Expression
Hypotheses	ceqsex2.1	|- F/ x ps
	ceqsex2.2	|- F/ y ch
	ceqsex2.3	|- A e. _V
	ceqsex2.4	|- B e. _V
	ceqsex2.5	|- ( x = A -> ( ph <-> ps ) )
	ceqsex2.6	|- ( y = B -> ( ps <-> ch ) )
Assertion	ceqsex2	|- ( E. x E. y ( x = A /\ y = B /\ ph ) <-> ch )

Proof

Step	Hyp	Ref	Expression
1		ceqsex2.1	|- F/ x ps
2		ceqsex2.2	|- F/ y ch
3		ceqsex2.3	|- A e. _V
4		ceqsex2.4	|- B e. _V
5		ceqsex2.5	|- ( x = A -> ( ph <-> ps ) )
6		ceqsex2.6	|- ( y = B -> ( ps <-> ch ) )
7		3anass	|- ( ( x = A /\ y = B /\ ph ) <-> ( x = A /\ ( y = B /\ ph ) ) )
8	7	exbii	|- ( E. y ( x = A /\ y = B /\ ph ) <-> E. y ( x = A /\ ( y = B /\ ph ) ) )
9		19.42v	|- ( E. y ( x = A /\ ( y = B /\ ph ) ) <-> ( x = A /\ E. y ( y = B /\ ph ) ) )
10	8 9	bitri	|- ( E. y ( x = A /\ y = B /\ ph ) <-> ( x = A /\ E. y ( y = B /\ ph ) ) )
11	10	exbii	|- ( E. x E. y ( x = A /\ y = B /\ ph ) <-> E. x ( x = A /\ E. y ( y = B /\ ph ) ) )
12		nfv	|- F/ x y = B
13	12 1	nfan	|- F/ x ( y = B /\ ps )
14	13	nfex	|- F/ x E. y ( y = B /\ ps )
15	5	anbi2d	|- ( x = A -> ( ( y = B /\ ph ) <-> ( y = B /\ ps ) ) )
16	15	exbidv	|- ( x = A -> ( E. y ( y = B /\ ph ) <-> E. y ( y = B /\ ps ) ) )
17	14 3 16	ceqsex	|- ( E. x ( x = A /\ E. y ( y = B /\ ph ) ) <-> E. y ( y = B /\ ps ) )
18	2 4 6	ceqsex	|- ( E. y ( y = B /\ ps ) <-> ch )
19	11 17 18	3bitri	|- ( E. x E. y ( x = A /\ y = B /\ ph ) <-> ch )