Theorem cdleme46frvlpq

Description: Show that ( FS ) is not under P .\/ Q when S isn't. (Contributed by NM, 1-Apr-2013)

	Ref	Expression
Hypotheses	cdlemef46.b	$$⊢B={\mathrm{Base}}_K$$
	cdlemef46.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	cdlemef46.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
	cdlemef46.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
	cdlemef46.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
	cdlemef46.h	$$⊢H=\mathrm{LHyp}\left(K\right)$$
	cdlemef46.u	$$⊢U=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }W$$
	cdlemef46.d	$$⊢D=\left(t\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
	cdlemefs46.e	$$⊢E=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(D\underset{\dot{}}{\vee }\left(\left(s\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
	cdlemef46.f	$$⊢F=\left(x\in B⟼if\left(\left(P\ne Q\wedge \neg x\underset{\dot{}}{\le }W\right),\left(\iota z\in B|\forall s\in A\left(\left(\neg s\underset{\dot{}}{\le }W\wedge s\underset{\dot{}}{\vee }\left(x\underset{\dot{}}{\wedge }W\right)=x\right)\to z=if\left(s\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right),\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=E\right)\right),⦋s/t⦌D\right)\underset{\dot{}}{\vee }\left(x\underset{\dot{}}{\wedge }W\right)\right)\right),x\right)\right)$$
Assertion	cdleme46frvlpq	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \left(S\in A\wedge \neg S\underset{\dot{}}{\le }W\right)\right)\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \neg F\left(S\right)\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$

Proof

Step	Hyp	Ref	Expression
1		cdlemef46.b	$$⊢B={\mathrm{Base}}_K$$
2		cdlemef46.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
3		cdlemef46.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
4		cdlemef46.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
5		cdlemef46.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
6		cdlemef46.h	$$⊢H=\mathrm{LHyp}\left(K\right)$$
7		cdlemef46.u	$$⊢U=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }W$$
8		cdlemef46.d	$$⊢D=\left(t\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
9		cdlemefs46.e	$$⊢E=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(D\underset{\dot{}}{\vee }\left(\left(s\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
10		cdlemef46.f	$$⊢F=\left(x\in B⟼if\left(\left(P\ne Q\wedge \neg x\underset{\dot{}}{\le }W\right),\left(\iota z\in B|\forall s\in A\left(\left(\neg s\underset{\dot{}}{\le }W\wedge s\underset{\dot{}}{\vee }\left(x\underset{\dot{}}{\wedge }W\right)=x\right)\to z=if\left(s\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right),\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=E\right)\right),⦋s/t⦌D\right)\underset{\dot{}}{\vee }\left(x\underset{\dot{}}{\wedge }W\right)\right)\right),x\right)\right)$$
11		eqid	$$⊢\left(S\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }S\right)\underset{\dot{}}{\wedge }W\right)\right)=\left(S\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }S\right)\underset{\dot{}}{\wedge }W\right)\right)$$
12	2 3 4 5 6 7 11	cdleme35fnpq	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \left(S\in A\wedge \neg S\underset{\dot{}}{\le }W\right)\right)\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \neg \left(\left(S\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }S\right)\underset{\dot{}}{\wedge }W\right)\right)\right)\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$
13	1 2 3 4 5 6 7 8 10	cdlemefr45e	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \left(S\in A\wedge \neg S\underset{\dot{}}{\le }W\right)\right)\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to F\left(S\right)=\left(S\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }S\right)\underset{\dot{}}{\wedge }W\right)\right)$$
14	13	breq1d	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \left(S\in A\wedge \neg S\underset{\dot{}}{\le }W\right)\right)\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \left(F\left(S\right)\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\leftrightarrow \left(\left(S\underset{\dot{}}{\vee }U\right)\underset{\dot{}}{\wedge }\left(Q\underset{\dot{}}{\vee }\left(\left(P\underset{\dot{}}{\vee }S\right)\underset{\dot{}}{\wedge }W\right)\right)\right)\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)$$
15	12 14	mtbird	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge W\in H\right)\wedge \left(P\in A\wedge \neg P\underset{\dot{}}{\le }W\right)\wedge \left(Q\in A\wedge \neg Q\underset{\dot{}}{\le }W\right)\right)\wedge \left(P\ne Q\wedge \left(S\in A\wedge \neg S\underset{\dot{}}{\le }W\right)\right)\wedge \neg S\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \neg F\left(S\right)\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)$$