Theorem cvlsupr3

Description: Two equivalent ways of expressing that R is a superposition of P and Q , which can replace the superposition part of ishlat1 , ( x =/= y -> E. z e. A ( z =/= x /\ z =/= y /\ z .<_ ( x .\/ y ) ) ) , with the simpler E. z e. A ( x .\/ z ) = ( y .\/ z ) as shown in ishlat3N . (Contributed by NM, 5-Nov-2012)

	Ref	Expression
Hypotheses	cvlsupr2.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
	cvlsupr2.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	cvlsupr2.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
Assertion	cvlsupr3	$$⊢\left(K\in \mathrm{CvLat}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\right)\to \left(P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\leftrightarrow \left(P\ne Q\to \left(R\ne P\wedge R\ne Q\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		cvlsupr2.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
2		cvlsupr2.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
3		cvlsupr2.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
4		df-ne	$$⊢P\ne Q\leftrightarrow \neg P=Q$$
5	4	imbi1i	$$⊢\left(P\ne Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\leftrightarrow \left(\neg P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)$$
6		oveq1	$$⊢P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R$$
7	6	biantrur	$$⊢\left(\neg P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\leftrightarrow \left(\left(P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\wedge \left(\neg P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\right)$$
8		pm4.83	$$⊢\left(\left(P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\wedge \left(\neg P=Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\right)\leftrightarrow P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R$$
9	5 7 8	3bitrri	$$⊢P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\leftrightarrow \left(P\ne Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)$$
10	1 2 3	cvlsupr2	$$⊢\left(K\in \mathrm{CvLat}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge P\ne Q\right)\to \left(P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\leftrightarrow \left(R\ne P\wedge R\ne Q\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)$$
11	10	3expia	$$⊢\left(K\in \mathrm{CvLat}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\right)\to \left(P\ne Q\to \left(P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\leftrightarrow \left(R\ne P\wedge R\ne Q\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\right)$$
12	11	pm5.74d	$$⊢\left(K\in \mathrm{CvLat}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\right)\to \left(\left(P\ne Q\to P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\right)\leftrightarrow \left(P\ne Q\to \left(R\ne P\wedge R\ne Q\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\right)$$
13	9 12	bitrid	$$⊢\left(K\in \mathrm{CvLat}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\right)\to \left(P\underset{\dot{}}{\vee }R=Q\underset{\dot{}}{\vee }R\leftrightarrow \left(P\ne Q\to \left(R\ne P\wedge R\ne Q\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\right)$$