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 ) )