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Metamath Proof Explorer

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 )