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