Metamath Proof Explorer

Theorem preq2d

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

Ref Expression
Hypothesis preq1d.1 
|- ( ph -> A = B )
Assertion preq2d 
|- ( ph -> { C , A } = { C , B } )

Proof

Step Hyp Ref Expression
1 preq1d.1 
 |-  ( ph -> A = B )
2 preq2 
 |-  ( A = B -> { C , A } = { C , B } )
3 1 2 syl 
 |-  ( ph -> { C , A } = { C , B } )