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	$$⊢\phi \to C\in \left(AFB\right)$$
	elimdelov.2	$$⊢Z\in \left(XFY\right)$$
Assertion	elimdelov	$$⊢if\left(\phi ,C,Z\right)\in \left(if\left(\phi ,A,X\right)Fif\left(\phi ,B,Y\right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		elimdelov.1	$$⊢\phi \to C\in \left(AFB\right)$$
2		elimdelov.2	$$⊢Z\in \left(XFY\right)$$
3		iftrue	$$⊢\phi \to if\left(\phi ,C,Z\right)=C$$
4		iftrue	$$⊢\phi \to if\left(\phi ,A,X\right)=A$$
5		iftrue	$$⊢\phi \to if\left(\phi ,B,Y\right)=B$$
6	4 5	oveq12d	$$⊢\phi \to if\left(\phi ,A,X\right)Fif\left(\phi ,B,Y\right)=AFB$$
7	1 3 6	3eltr4d	$$⊢\phi \to if\left(\phi ,C,Z\right)\in \left(if\left(\phi ,A,X\right)Fif\left(\phi ,B,Y\right)\right)$$
8		iffalse	$$⊢\neg \phi \to if\left(\phi ,C,Z\right)=Z$$
9	8 2	eqeltrdi	$$⊢\neg \phi \to if\left(\phi ,C,Z\right)\in \left(XFY\right)$$
10		iffalse	$$⊢\neg \phi \to if\left(\phi ,A,X\right)=X$$
11		iffalse	$$⊢\neg \phi \to if\left(\phi ,B,Y\right)=Y$$
12	10 11	oveq12d	$$⊢\neg \phi \to if\left(\phi ,A,X\right)Fif\left(\phi ,B,Y\right)=XFY$$
13	9 12	eleqtrrd	$$⊢\neg \phi \to if\left(\phi ,C,Z\right)\in \left(if\left(\phi ,A,X\right)Fif\left(\phi ,B,Y\right)\right)$$
14	7 13	pm2.61i	$$⊢if\left(\phi ,C,Z\right)\in \left(if\left(\phi ,A,X\right)Fif\left(\phi ,B,Y\right)\right)$$