Metamath Proof Explorer


Theorem hpgne2

Description: Points on the open half plane cannot lie on its border. (Contributed by Thierry Arnoux, 4-Mar-2020)

Ref Expression
Hypotheses ishpg.p ⊢ P = Base G
ishpg.i ⊢ I = Itv ⁡ G
ishpg.l ⊢ L = Line 𝒢 ⁡ G
ishpg.o ⊢ O = a b | a ∈ P ∖ D ∧ b ∈ P ∖ D ∧ ∃ t ∈ D t ∈ a I b
ishpg.g ⊢ φ → G ∈ 𝒢 Tarski
ishpg.d ⊢ φ → D ∈ ran ⁡ L
hpgbr.a ⊢ φ → A ∈ P
hpgbr.b ⊢ φ → B ∈ P
hpgne1.1 ⊢ φ → A hp 𝒢 ⁡ G ⁡ D B
Assertion hpgne2 ⊢ φ → ¬ B ∈ D

Proof

Step Hyp Ref Expression
1 ishpg.p ⊢ P = Base G
2 ishpg.i ⊢ I = Itv ⁡ G
3 ishpg.l ⊢ L = Line 𝒢 ⁡ G
4 ishpg.o ⊢ O = a b | a ∈ P ∖ D ∧ b ∈ P ∖ D ∧ ∃ t ∈ D t ∈ a I b
5 ishpg.g ⊢ φ → G ∈ 𝒢 Tarski
6 ishpg.d ⊢ φ → D ∈ ran ⁡ L
7 hpgbr.a ⊢ φ → A ∈ P
8 hpgbr.b ⊢ φ → B ∈ P
9 hpgne1.1 ⊢ φ → A hp 𝒢 ⁡ G ⁡ D B
10 eqid ⊢ dist ⁡ G = dist ⁡ G
11 6 ad2antrr ⊢ φ ∧ c ∈ P ∧ A O c ∧ B O c → D ∈ ran ⁡ L
12 5 ad2antrr ⊢ φ ∧ c ∈ P ∧ A O c ∧ B O c → G ∈ 𝒢 Tarski
13 8 ad2antrr ⊢ φ ∧ c ∈ P ∧ A O c ∧ B O c → B ∈ P
14 simplr ⊢ φ ∧ c ∈ P ∧ A O c ∧ B O c → c ∈ P
15 simprr ⊢ φ ∧ c ∈ P ∧ A O c ∧ B O c → B O c
16 1 10 2 4 3 11 12 13 14 15 oppne1 ⊢ φ ∧ c ∈ P ∧ A O c ∧ B O c → ¬ B ∈ D
17 1 2 3 4 5 6 7 8 hpgbr ⊢ φ → A hp 𝒢 ⁡ G ⁡ D B ↔ ∃ c ∈ P A O c ∧ B O c
18 9 17 mpbid ⊢ φ → ∃ c ∈ P A O c ∧ B O c
19 16 18 r19.29a ⊢ φ → ¬ B ∈ D