Theorem dfif6

Description: An alternate definition of the conditional operator df-if as a simple class abstraction. (Contributed by Mario Carneiro, 8-Sep-2013)

	Ref	Expression
Assertion	dfif6	$$⊢if\left(\phi ,A,B\right)=\left\{x\in A|\phi \right\}\cup \left\{x\in B|\neg \phi \right\}$$

Proof

Step	Hyp	Ref	Expression
1		eleq1w	$$⊢x=y\to \left(x\in A\leftrightarrow y\in A\right)$$
2	1	anbi1d	$$⊢x=y\to \left(\left(x\in A\wedge \phi \right)\leftrightarrow \left(y\in A\wedge \phi \right)\right)$$
3		eleq1w	$$⊢x=y\to \left(x\in B\leftrightarrow y\in B\right)$$
4	3	anbi1d	$$⊢x=y\to \left(\left(x\in B\wedge \neg \phi \right)\leftrightarrow \left(y\in B\wedge \neg \phi \right)\right)$$
5	2 4	unabw	$$⊢\left\{x|\left(x\in A\wedge \phi \right)\right\}\cup \left\{x|\left(x\in B\wedge \neg \phi \right)\right\}=\left\{y|\left(\left(y\in A\wedge \phi \right)\vee \left(y\in B\wedge \neg \phi \right)\right)\right\}$$
6		df-rab	$$⊢\left\{x\in A|\phi \right\}=\left\{x|\left(x\in A\wedge \phi \right)\right\}$$
7		df-rab	$$⊢\left\{x\in B|\neg \phi \right\}=\left\{x|\left(x\in B\wedge \neg \phi \right)\right\}$$
8	6 7	uneq12i	$$⊢\left\{x\in A|\phi \right\}\cup \left\{x\in B|\neg \phi \right\}=\left\{x|\left(x\in A\wedge \phi \right)\right\}\cup \left\{x|\left(x\in B\wedge \neg \phi \right)\right\}$$
9		df-if	$$⊢if\left(\phi ,A,B\right)=\left\{y|\left(\left(y\in A\wedge \phi \right)\vee \left(y\in B\wedge \neg \phi \right)\right)\right\}$$
10	5 8 9	3eqtr4ri	$$⊢if\left(\phi ,A,B\right)=\left\{x\in A|\phi \right\}\cup \left\{x\in B|\neg \phi \right\}$$