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 = Base G
legval.d - ˙ = dist G
legval.i I = Itv G
legval.l ˙ = 𝒢 G
legval.g φ G 𝒢 Tarski
legid.a φ A P
legid.b φ B P
legtrd.c φ C P
legtrd.d φ D P
legeq.1 φ A - ˙ B ˙ C - ˙ C
Assertion legeq φ A = B

Proof

Step Hyp Ref Expression
1 legval.p P = Base G
2 legval.d - ˙ = dist G
3 legval.i I = Itv G
4 legval.l ˙ = 𝒢 G
5 legval.g φ G 𝒢 Tarski
6 legid.a φ A P
7 legid.b φ B P
8 legtrd.c φ C P
9 legtrd.d φ D P
10 legeq.1 φ A - ˙ B ˙ C - ˙ C
11 1 2 3 4 5 8 6 6 7 leg0 φ C - ˙ C ˙ A - ˙ B
12 1 2 3 4 5 6 7 8 8 10 11 legtri3 φ A - ˙ B = C - ˙ C
13 1 2 3 5 6 7 8 12 axtgcgrid φ A = B