Metamath Proof Explorer


Theorem preq12d

Description: Equality deduction for unordered pairs. (Contributed by NM, 19-Oct-2012)

Ref Expression
Hypotheses preq1d.1
|- ( ph -> A = B )
preq12d.2
|- ( ph -> C = D )
Assertion preq12d
|- ( ph -> { A , C } = { B , D } )

Proof

Step Hyp Ref Expression
1 preq1d.1
 |-  ( ph -> A = B )
2 preq12d.2
 |-  ( ph -> C = D )
3 preq12
 |-  ( ( A = B /\ C = D ) -> { A , C } = { B , D } )
4 1 2 3 syl2anc
 |-  ( ph -> { A , C } = { B , D } )