Metamath Proof Explorer


Theorem ifeq1

Description: Equality theorem for conditional operator. (Contributed by NM, 1-Sep-2004) (Revised by Mario Carneiro, 8-Sep-2013)

Ref Expression
Assertion ifeq1
|- ( A = B -> if ( ph , A , C ) = if ( ph , B , C ) )

Proof

Step Hyp Ref Expression
1 rabeq
 |-  ( A = B -> { x e. A | ph } = { x e. B | ph } )
2 1 uneq1d
 |-  ( A = B -> ( { x e. A | ph } u. { x e. C | -. ph } ) = ( { x e. B | ph } u. { x e. C | -. ph } ) )
3 dfif6
 |-  if ( ph , A , C ) = ( { x e. A | ph } u. { x e. C | -. ph } )
4 dfif6
 |-  if ( ph , B , C ) = ( { x e. B | ph } u. { x e. C | -. ph } )
5 2 3 4 3eqtr4g
 |-  ( A = B -> if ( ph , A , C ) = if ( ph , B , C ) )