Theorem dath

Description: Desargues's theorem of projective geometry (proved for a Hilbert lattice). Assume each triple of atoms (points) P Q R and S T U forms a triangle (i.e. determines a plane). Assume that lines P S , Q T , and R U meet at a "center of perspectivity" C . (We also assume that C is not on any of the 6 lines forming the two triangles.) Then the atoms D = ( P .\/ Q ) ./\ ( S .\/ T ) , E = ( Q .\/ R ) ./\ ( T .\/ U ) , F = ( R .\/ P ) ./\ ( U .\/ S ) are colinear, forming an "axis of perspectivity".

Our proof roughly follows Theorem 2.7.1, p. 78 in Beutelspacher and Rosenbaum,Projective Geometry: From Foundations to Applications, Cambridge University Press (1988). Unlike them, we do not assume that C is an atom to make this theorem slightly more general for easier future use. However, we prove that C must be an atom in dalemcea .

For a visual demonstration, see the "Desargues's theorem" applet at http://www.dynamicgeometry.com/JavaSketchpad/Gallery.html . The points I, J, and K there define the axis of perspectivity.

See Theorems dalaw for Desargues's law, which eliminates all of the preconditions on the atoms except for central perspectivity. This is Metamath 100 proof #87. (Contributed by NM, 20-Aug-2012)

	Ref	Expression
Hypotheses	dath.b	$$⊢B={\mathrm{Base}}_K$$
	dath.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	dath.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
	dath.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
	dath.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
	dath.o	$$⊢O=\mathrm{LPlanes}\left(K\right)$$
	dath.d	$$⊢D=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(S\underset{\dot{}}{\vee }T\right)$$
	dath.e	$$⊢E=\left(Q\underset{\dot{}}{\vee }R\right)\underset{\dot{}}{\wedge }\left(T\underset{\dot{}}{\vee }U\right)$$
	dath.f	$$⊢F=\left(R\underset{\dot{}}{\vee }P\right)\underset{\dot{}}{\wedge }\left(U\underset{\dot{}}{\vee }S\right)$$
Assertion	dath	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge C\in B\right)\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \left(S\in A\wedge T\in A\wedge U\in A\right)\right)\wedge \left(\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\vee }R\in O\wedge \left(S\underset{\dot{}}{\vee }T\right)\underset{\dot{}}{\vee }U\in O\right)\wedge \left(\left(\neg C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\wedge \neg C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }R\right)\wedge \neg C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }P\right)\right)\wedge \left(\neg C\underset{\dot{}}{\le }\left(S\underset{\dot{}}{\vee }T\right)\wedge \neg C\underset{\dot{}}{\le }\left(T\underset{\dot{}}{\vee }U\right)\wedge \neg C\underset{\dot{}}{\le }\left(U\underset{\dot{}}{\vee }S\right)\right)\wedge \left(C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }S\right)\wedge C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }T\right)\wedge C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }U\right)\right)\right)\right)\to F\underset{\dot{}}{\le }\left(D\underset{\dot{}}{\vee }E\right)$$

Proof

Step	Hyp	Ref	Expression
1		dath.b	$$⊢B={\mathrm{Base}}_K$$
2		dath.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
3		dath.j	$$⊢\underset{\dot{}}{\vee }=\mathrm{join}\left(K\right)$$
4		dath.a	$$⊢A=\mathrm{Atoms}\left(K\right)$$
5		dath.m	$$⊢\underset{\dot{}}{\wedge }=\mathrm{meet}\left(K\right)$$
6		dath.o	$$⊢O=\mathrm{LPlanes}\left(K\right)$$
7		dath.d	$$⊢D=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(S\underset{\dot{}}{\vee }T\right)$$
8		dath.e	$$⊢E=\left(Q\underset{\dot{}}{\vee }R\right)\underset{\dot{}}{\wedge }\left(T\underset{\dot{}}{\vee }U\right)$$
9		dath.f	$$⊢F=\left(R\underset{\dot{}}{\vee }P\right)\underset{\dot{}}{\wedge }\left(U\underset{\dot{}}{\vee }S\right)$$
10	1	eleq2i	$$⊢C\in B\leftrightarrow C\in {\mathrm{Base}}_K$$
11	10	anbi2i	$$⊢\left(K\in \mathrm{HL}\wedge C\in B\right)\leftrightarrow \left(K\in \mathrm{HL}\wedge C\in {\mathrm{Base}}_K\right)$$
12	11	3anbi1i	$$⊢\left(\left(K\in \mathrm{HL}\wedge C\in B\right)\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \left(S\in A\wedge T\in A\wedge U\in A\right)\right)\leftrightarrow \left(\left(K\in \mathrm{HL}\wedge C\in {\mathrm{Base}}_K\right)\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \left(S\in A\wedge T\in A\wedge U\in A\right)\right)$$
13	12	3anbi1i	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge C\in B\right)\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \left(S\in A\wedge T\in A\wedge U\in A\right)\right)\wedge \left(\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\vee }R\in O\wedge \left(S\underset{\dot{}}{\vee }T\right)\underset{\dot{}}{\vee }U\in O\right)\wedge \left(\left(\neg C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\wedge \neg C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }R\right)\wedge \neg C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }P\right)\right)\wedge \left(\neg C\underset{\dot{}}{\le }\left(S\underset{\dot{}}{\vee }T\right)\wedge \neg C\underset{\dot{}}{\le }\left(T\underset{\dot{}}{\vee }U\right)\wedge \neg C\underset{\dot{}}{\le }\left(U\underset{\dot{}}{\vee }S\right)\right)\wedge \left(C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }S\right)\wedge C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }T\right)\wedge C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }U\right)\right)\right)\right)\leftrightarrow \left(\left(\left(K\in \mathrm{HL}\wedge C\in {\mathrm{Base}}_K\right)\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \left(S\in A\wedge T\in A\wedge U\in A\right)\right)\wedge \left(\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\vee }R\in O\wedge \left(S\underset{\dot{}}{\vee }T\right)\underset{\dot{}}{\vee }U\in O\right)\wedge \left(\left(\neg C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\wedge \neg C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }R\right)\wedge \neg C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }P\right)\right)\wedge \left(\neg C\underset{\dot{}}{\le }\left(S\underset{\dot{}}{\vee }T\right)\wedge \neg C\underset{\dot{}}{\le }\left(T\underset{\dot{}}{\vee }U\right)\wedge \neg C\underset{\dot{}}{\le }\left(U\underset{\dot{}}{\vee }S\right)\right)\wedge \left(C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }S\right)\wedge C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }T\right)\wedge C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }U\right)\right)\right)\right)$$
14		eqid	$$⊢\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\vee }R=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\vee }R$$
15		eqid	$$⊢\left(S\underset{\dot{}}{\vee }T\right)\underset{\dot{}}{\vee }U=\left(S\underset{\dot{}}{\vee }T\right)\underset{\dot{}}{\vee }U$$
16	13 2 3 4 5 6 14 15 7 8 9	dalem63	$$⊢\left(\left(\left(K\in \mathrm{HL}\wedge C\in B\right)\wedge \left(P\in A\wedge Q\in A\wedge R\in A\right)\wedge \left(S\in A\wedge T\in A\wedge U\in A\right)\right)\wedge \left(\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\vee }R\in O\wedge \left(S\underset{\dot{}}{\vee }T\right)\underset{\dot{}}{\vee }U\in O\right)\wedge \left(\left(\neg C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\wedge \neg C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }R\right)\wedge \neg C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }P\right)\right)\wedge \left(\neg C\underset{\dot{}}{\le }\left(S\underset{\dot{}}{\vee }T\right)\wedge \neg C\underset{\dot{}}{\le }\left(T\underset{\dot{}}{\vee }U\right)\wedge \neg C\underset{\dot{}}{\le }\left(U\underset{\dot{}}{\vee }S\right)\right)\wedge \left(C\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }S\right)\wedge C\underset{\dot{}}{\le }\left(Q\underset{\dot{}}{\vee }T\right)\wedge C\underset{\dot{}}{\le }\left(R\underset{\dot{}}{\vee }U\right)\right)\right)\right)\to F\underset{\dot{}}{\le }\left(D\underset{\dot{}}{\vee }E\right)$$