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Metamath Proof Explorer

Theorem elimdelov

Description: Eliminate a hypothesis which is a predicate expressing membership in the result of an operator (deduction version). (Contributed by Paul Chapman, 25-Mar-2008)

Ref Expression
Hypotheses elimdelov.1 
|- ( ph -> C e. ( A F B ) )
elimdelov.2 
|- Z e. ( X F Y )
Assertion elimdelov 
|- if ( ph , C , Z ) e. ( if ( ph , A , X ) F if ( ph , B , Y ) )

Proof

Step Hyp Ref Expression
1 elimdelov.1 
 |-  ( ph -> C e. ( A F B ) )
2 elimdelov.2 
 |-  Z e. ( X F Y )
3 iftrue 
 |-  ( ph -> if ( ph , C , Z ) = C )
4 iftrue 
 |-  ( ph -> if ( ph , A , X ) = A )
5 iftrue 
 |-  ( ph -> if ( ph , B , Y ) = B )
6 4 5 oveq12d 
 |-  ( ph -> ( if ( ph , A , X ) F if ( ph , B , Y ) ) = ( A F B ) )
7 1 3 6 3eltr4d 
 |-  ( ph -> if ( ph , C , Z ) e. ( if ( ph , A , X ) F if ( ph , B , Y ) ) )
8 iffalse 
 |-  ( -. ph -> if ( ph , C , Z ) = Z )
9 8 2 eqeltrdi 
 |-  ( -. ph -> if ( ph , C , Z ) e. ( X F Y ) )
10 iffalse 
 |-  ( -. ph -> if ( ph , A , X ) = X )
11 iffalse 
 |-  ( -. ph -> if ( ph , B , Y ) = Y )
12 10 11 oveq12d 
 |-  ( -. ph -> ( if ( ph , A , X ) F if ( ph , B , Y ) ) = ( X F Y ) )
13 9 12 eleqtrrd 
 |-  ( -. ph -> if ( ph , C , Z ) e. ( if ( ph , A , X ) F if ( ph , B , Y ) ) )
14 7 13 pm2.61i 
 |-  if ( ph , C , Z ) e. ( if ( ph , A , X ) F if ( ph , B , Y ) )