Metamath Proof Explorer


Theorem legeq

Description: Deduce equality from "less than" null segments. (Contributed by Thierry Arnoux, 12-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
legeq.1 φA-˙B˙C-˙C
Assertion legeq φA=B

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 legeq.1 φA-˙B˙C-˙C
11 1 2 3 4 5 8 6 6 7 leg0 φC-˙C˙A-˙B
12 1 2 3 4 5 6 7 8 8 10 11 legtri3 φA-˙B=C-˙C
13 1 2 3 5 6 7 8 12 axtgcgrid φA=B