Metamath Proof Explorer

Theorem preq1d

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

Ref Expression
Hypothesis preq1d.1 
|- ( ph -> A = B )
Assertion preq1d 
|- ( ph -> { A , C } = { B , C } )

Proof

Step Hyp Ref Expression
1 preq1d.1 
 |-  ( ph -> A = B )
2 preq1 
 |-  ( A = B -> { A , C } = { B , C } )
3 1 2 syl 
 |-  ( ph -> { A , C } = { B , C } )