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Metamath Proof Explorer

Theorem eqopab2b

Description: Equivalence of ordered pair abstraction equality and biconditional. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker eqopab2bw when possible. (Contributed by Mario Carneiro, 4-Jan-2017) (New usage is discouraged.)

Ref Expression
Assertion eqopab2b 
|- ( { <. x , y >. | ph } = { <. x , y >. | ps } <-> A. x A. y ( ph <-> ps ) )

Proof

Step Hyp Ref Expression
1 ssopab2b 
 |-  ( { <. x , y >. | ph } C_ { <. x , y >. | ps } <-> A. x A. y ( ph -> ps ) )
2 ssopab2b 
 |-  ( { <. x , y >. | ps } C_ { <. x , y >. | ph } <-> A. x A. y ( ps -> ph ) )
3 1 2 anbi12i 
 |-  ( ( { <. x , y >. | ph } C_ { <. x , y >. | ps } /\ { <. x , y >. | ps } C_ { <. x , y >. | ph } ) <-> ( A. x A. y ( ph -> ps ) /\ A. x A. y ( ps -> ph ) ) )
4 eqss 
 |-  ( { <. x , y >. | ph } = { <. x , y >. | ps } <-> ( { <. x , y >. | ph } C_ { <. x , y >. | ps } /\ { <. x , y >. | ps } C_ { <. x , y >. | ph } ) )
5 2albiim 
 |-  ( A. x A. y ( ph <-> ps ) <-> ( A. x A. y ( ph -> ps ) /\ A. x A. y ( ps -> ph ) ) )
6 3 4 5 3bitr4i 
 |-  ( { <. x , y >. | ph } = { <. x , y >. | ps } <-> A. x A. y ( ph <-> ps ) )