Metamath Proof Explorer


Theorem legbtwn

Description: Deduce betweenness from "less than" relation. Corresponds loosely to Proposition 6.13 of Schwabhauser p. 45. (Contributed by Thierry Arnoux, 25-Aug-2019)

Ref Expression
Hypotheses legval.p P=BaseG
legval.d -˙=distG
legval.i I=ItvG
legval.l ˙=𝒢G
legval.g φG𝒢Tarski
legid.a φAP
legid.b φBP
legtrd.c φCP
legtrd.d φDP
legbtwn.1 φACIBBCIA
legbtwn.2 φC-˙A˙C-˙B
Assertion legbtwn φACIB

Proof

Step Hyp Ref Expression
1 legval.p P=BaseG
2 legval.d -˙=distG
3 legval.i I=ItvG
4 legval.l ˙=𝒢G
5 legval.g φG𝒢Tarski
6 legid.a φAP
7 legid.b φBP
8 legtrd.c φCP
9 legtrd.d φDP
10 legbtwn.1 φACIBBCIA
11 legbtwn.2 φC-˙A˙C-˙B
12 simpr φACIBACIB
13 5 adantr φBCIAG𝒢Tarski
14 6 adantr φBCIAAP
15 7 adantr φBCIABP
16 8 adantr φBCIACP
17 simpr φBCIABCIA
18 1 2 3 13 16 15 14 17 tgbtwncom φBCIABAIC
19 1 2 3 13 15 16 tgbtwntriv1 φBCIABBIC
20 11 adantr φBCIAC-˙A˙C-˙B
21 1 2 3 4 13 16 15 14 17 btwnleg φBCIAC-˙B˙C-˙A
22 1 2 3 4 13 16 14 16 15 20 21 legtri3 φBCIAC-˙A=C-˙B
23 1 2 3 13 16 14 16 15 22 tgcgrcomlr φBCIAA-˙C=B-˙C
24 eqidd φBCIAB-˙C=B-˙C
25 1 2 3 13 14 15 16 15 15 16 18 19 23 24 tgcgrsub φBCIAA-˙B=B-˙B
26 1 2 3 13 14 15 15 25 axtgcgrid φBCIAA=B
27 26 17 eqeltrd φBCIAACIA
28 26 oveq2d φBCIACIA=CIB
29 27 28 eleqtrd φBCIAACIB
30 12 29 10 mpjaodan φACIB