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Metamath Proof Explorer

Theorem ibllem

Description: Conditioned equality theorem for the if statement. (Contributed by Mario Carneiro, 31-Jul-2014)

Ref Expression
Hypothesis ibllem.1 
|- ( ( ph /\ x e. A ) -> B = C )
Assertion ibllem 
|- ( ph -> if ( ( x e. A /\ 0 <_ B ) , B , 0 ) = if ( ( x e. A /\ 0 <_ C ) , C , 0 ) )

Proof

Step Hyp Ref Expression
1 ibllem.1 
 |-  ( ( ph /\ x e. A ) -> B = C )
2 1 breq2d 
 |-  ( ( ph /\ x e. A ) -> ( 0 <_ B <-> 0 <_ C ) )
3 2 pm5.32da 
 |-  ( ph -> ( ( x e. A /\ 0 <_ B ) <-> ( x e. A /\ 0 <_ C ) ) )
4 3 ifbid 
 |-  ( ph -> if ( ( x e. A /\ 0 <_ B ) , B , 0 ) = if ( ( x e. A /\ 0 <_ C ) , B , 0 ) )
5 1 adantrr 
 |-  ( ( ph /\ ( x e. A /\ 0 <_ C ) ) -> B = C )
6 5 ifeq1da 
 |-  ( ph -> if ( ( x e. A /\ 0 <_ C ) , B , 0 ) = if ( ( x e. A /\ 0 <_ C ) , C , 0 ) )
7 4 6 eqtrd 
 |-  ( ph -> if ( ( x e. A /\ 0 <_ B ) , B , 0 ) = if ( ( x e. A /\ 0 <_ C ) , C , 0 ) )