Theorem eldifpr

Description: Membership in a set with two elements removed. Similar to eldifsn and eldiftp . (Contributed by Mario Carneiro, 18-Jul-2017)

	Ref	Expression
Assertion	eldifpr	$$⊢A\in \left(B\setminus \left\{C,D\right\}\right)\leftrightarrow \left(A\in B\wedge A\ne C\wedge A\ne D\right)$$

Proof

Step	Hyp	Ref	Expression
1		elprg	$$⊢A\in B\to \left(A\in \left\{C,D\right\}\leftrightarrow \left(A=C\vee A=D\right)\right)$$
2	1	notbid	$$⊢A\in B\to \left(\neg A\in \left\{C,D\right\}\leftrightarrow \neg \left(A=C\vee A=D\right)\right)$$
3		neanior	$$⊢\left(A\ne C\wedge A\ne D\right)\leftrightarrow \neg \left(A=C\vee A=D\right)$$
4	2 3	syl6bbr	$$⊢A\in B\to \left(\neg A\in \left\{C,D\right\}\leftrightarrow \left(A\ne C\wedge A\ne D\right)\right)$$
5	4	pm5.32i	$$⊢\left(A\in B\wedge \neg A\in \left\{C,D\right\}\right)\leftrightarrow \left(A\in B\wedge \left(A\ne C\wedge A\ne D\right)\right)$$
6		eldif	$$⊢A\in \left(B\setminus \left\{C,D\right\}\right)\leftrightarrow \left(A\in B\wedge \neg A\in \left\{C,D\right\}\right)$$
7		3anass	$$⊢\left(A\in B\wedge A\ne C\wedge A\ne D\right)\leftrightarrow \left(A\in B\wedge \left(A\ne C\wedge A\ne D\right)\right)$$
8	5 6 7	3bitr4i	$$⊢A\in \left(B\setminus \left\{C,D\right\}\right)\leftrightarrow \left(A\in B\wedge A\ne C\wedge A\ne D\right)$$