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Metamath Proof Explorer

Theorem elpreqpr

Description: Equality and membership rule for pairs. (Contributed by Scott Fenton, 7-Dec-2020)

Ref Expression
Assertion elpreqpr 
|- ( A e. { B , C } -> E. x { B , C } = { A , x } )

Proof

Step Hyp Ref Expression
1 elpri 
 |-  ( A e. { B , C } -> ( A = B \/ A = C ) )
2 elex 
 |-  ( A e. { B , C } -> A e. _V )
3 elpreqprlem 
 |-  ( B e. _V -> E. x { B , C } = { B , x } )
4 eleq1 
 |-  ( A = B -> ( A e. _V <-> B e. _V ) )
5 preq1 
 |-  ( A = B -> { A , x } = { B , x } )
6 5 eqeq2d 
 |-  ( A = B -> ( { B , C } = { A , x } <-> { B , C } = { B , x } ) )
7 6 exbidv 
 |-  ( A = B -> ( E. x { B , C } = { A , x } <-> E. x { B , C } = { B , x } ) )
8 4 7 imbi12d 
 |-  ( A = B -> ( ( A e. _V -> E. x { B , C } = { A , x } ) <-> ( B e. _V -> E. x { B , C } = { B , x } ) ) )
9 3 8 mpbiri 
 |-  ( A = B -> ( A e. _V -> E. x { B , C } = { A , x } ) )
10 9 imp 
 |-  ( ( A = B /\ A e. _V ) -> E. x { B , C } = { A , x } )
11 elpreqprlem 
 |-  ( C e. _V -> E. x { C , B } = { C , x } )
12 prcom 
 |-  { C , B } = { B , C }
13 12 eqeq1i 
 |-  ( { C , B } = { C , x } <-> { B , C } = { C , x } )
14 13 exbii 
 |-  ( E. x { C , B } = { C , x } <-> E. x { B , C } = { C , x } )
15 11 14 sylib 
 |-  ( C e. _V -> E. x { B , C } = { C , x } )
16 eleq1 
 |-  ( A = C -> ( A e. _V <-> C e. _V ) )
17 preq1 
 |-  ( A = C -> { A , x } = { C , x } )
18 17 eqeq2d 
 |-  ( A = C -> ( { B , C } = { A , x } <-> { B , C } = { C , x } ) )
19 18 exbidv 
 |-  ( A = C -> ( E. x { B , C } = { A , x } <-> E. x { B , C } = { C , x } ) )
20 16 19 imbi12d 
 |-  ( A = C -> ( ( A e. _V -> E. x { B , C } = { A , x } ) <-> ( C e. _V -> E. x { B , C } = { C , x } ) ) )
21 15 20 mpbiri 
 |-  ( A = C -> ( A e. _V -> E. x { B , C } = { A , x } ) )
22 21 imp 
 |-  ( ( A = C /\ A e. _V ) -> E. x { B , C } = { A , x } )
23 10 22 jaoian 
 |-  ( ( ( A = B \/ A = C ) /\ A e. _V ) -> E. x { B , C } = { A , x } )
24 1 2 23 syl2anc 
 |-  ( A e. { B , C } -> E. x { B , C } = { A , x } )