Theorem cdleme00a

Description: Part of proof of Lemma E in Crawley p. 113. (Contributed by NM, 14-Jun-2012)

	Ref	Expression
Hypotheses	cdleme0.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	cdleme0.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
	cdleme0.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
	cdleme0.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
Assertion	cdleme00a	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to R\ne P$$

Proof

Step	Hyp	Ref	Expression
1		cdleme0.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
2		cdleme0.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
3		cdleme0.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
4		cdleme0.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
5		simp1	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to K\in \mathrm{HL}$$
6		simp23	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to R\in A$$
7		simp21	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to P\in A$$
8		simp22	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to Q\in A$$
9		simp3	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$
10	1 2 4	atnlej1	$$⊢\left(K\in \mathrm{HL}\wedge \left(R\in A\wedge P\in A\wedge Q\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to R\ne P$$
11	5 6 7 8 9 10	syl131anc	$$⊢\left(K\in \mathrm{HL}\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \neg R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to R\ne P$$