Theorem cdleme3fN

Description: Part of proof of Lemma E in Crawley p. 113. Lemma leading to cdleme3fa and cdleme3 . TODO: Delete - duplicates cdleme0e . (Contributed by NM, 6-Jun-2012) (New usage is discouraged.)

	Ref	Expression
Hypotheses	cdleme1.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	cdleme1.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
	cdleme1.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
	cdleme1.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
	cdleme1.h	$$⊢H=\mathrm{LHyp}\left(K\right)$$
	cdleme1.u	$$⊢U=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }W$$
	cdleme1.f	$$⊢F=\left(R\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }R\right)\underset{\dot{}}{\wedge }W\right)\right)$$
	cdleme3.3	$$⊢V=\left(P\underset{\dot{}}{\vee }R\right)\underset{\dot{}}{\wedge }W$$
Assertion	cdleme3fN	$$⊢\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(\left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge Q\in A\wedge \left(R\in A\wedge \neg R\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\to U\ne V$$

Proof

Step	Hyp	Ref	Expression
1		cdleme1.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
2		cdleme1.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
3		cdleme1.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
4		cdleme1.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
5		cdleme1.h	$$⊢H=\mathrm{LHyp}\left(K\right)$$
6		cdleme1.u	$$⊢U=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }W$$
7		cdleme1.f	$$⊢F=\left(R\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }R\right)\underset{\dot{}}{\wedge }W\right)\right)$$
8		cdleme3.3	$$⊢V=\left(P\underset{\dot{}}{\vee }R\right)\underset{\dot{}}{\wedge }W$$
9	1 2 3 4 5 6 8	cdleme0e	$$⊢\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(\left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge Q\in A\wedge \left(R\in A\wedge \neg R\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\right)\to U\ne V$$