Theorem dihmeetlem5

Description: Part of proof that isomorphism H is order-preserving . (Contributed by NM, 6-Apr-2014)

	Ref	Expression
Hypotheses	dihmeetlem5.b	$$⊢B={\mathrm{Base}}_K$$
	dihmeetlem5.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	dihmeetlem5.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
	dihmeetlem5.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
	dihmeetlem5.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
Assertion	dihmeetlem5	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to X\underset{\dot{}}{\wedge }\left(Y\underset{\dot{}}{\vee }Q\right)=\left(X\underset{\dot{}}{\wedge }Y\right)\underset{\dot{}}{\vee }Q$$

Proof

Step	Hyp	Ref	Expression
1		dihmeetlem5.b	$$⊢B={\mathrm{Base}}_K$$
2		dihmeetlem5.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
3		dihmeetlem5.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
4		dihmeetlem5.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
5		dihmeetlem5.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
6		simpl1	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to K\in \mathrm{HL}$$
7		simprl	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to Q\in A$$
8		simpl2	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to X\in B$$
9		simpl3	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to Y\in B$$
10		simprr	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to Q\underset{\dot{}}{\le }X$$
11	1 2 3 4 5	atmod2i1	$$⊢\left(K\in \mathrm{HL}\wedge \left(Q\in A\wedge X\in B\wedge Y\in B\right)\wedge Q\underset{\dot{}}{\le }X\right)\to \left(X\underset{\dot{}}{\wedge }Y\right)\underset{\dot{}}{\vee }Q=X\underset{\dot{}}{\wedge }\left(Y\underset{\dot{}}{\vee }Q\right)$$
12	6 7 8 9 10 11	syl131anc	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to \left(X\underset{\dot{}}{\wedge }Y\right)\underset{\dot{}}{\vee }Q=X\underset{\dot{}}{\wedge }\left(Y\underset{\dot{}}{\vee }Q\right)$$
13	12	eqcomd	$$⊢\left(\left(K\in \mathrm{HL}\wedge X\in B\wedge Y\in B\right)\wedge \left(Q\in A\wedge Q\underset{\dot{}}{\le }X\right)\right)\to X\underset{\dot{}}{\wedge }\left(Y\underset{\dot{}}{\vee }Q\right)=\left(X\underset{\dot{}}{\wedge }Y\right)\underset{\dot{}}{\vee }Q$$