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Metamath Proof Explorer

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 )