Metamath Proof Explorer


Theorem inagflat

Description: Any point lies in a flat angle. (Contributed by Thierry Arnoux, 13-Feb-2023)

Ref Expression
Hypotheses isinag.p ⊢ P = Base G
isinag.i ⊢ I = Itv ⁡ G
isinag.k ⊢ K = hl 𝒢 ⁡ G
isinag.x ⊢ φ → X ∈ P
isinag.a ⊢ φ → A ∈ P
isinag.b ⊢ φ → B ∈ P
isinag.c ⊢ φ → C ∈ P
inagflat.g ⊢ φ → G ∈ 𝒢 Tarski
inagflat.x ⊢ φ → X ∈ P
inagflat.1 ⊢ φ → A ≠ B
inagflat.2 ⊢ φ → C ≠ B
inagflat.3 ⊢ φ → X ≠ B
inagflat.4 ⊢ φ → B ∈ A I C
Assertion inagflat ⊢ φ → X ∈ 𝒢 ∠ ⁡ G ⟨“ ABC ”⟩

Proof

Step Hyp Ref Expression
1 isinag.p ⊢ P = Base G
2 isinag.i ⊢ I = Itv ⁡ G
3 isinag.k ⊢ K = hl 𝒢 ⁡ G
4 isinag.x ⊢ φ → X ∈ P
5 isinag.a ⊢ φ → A ∈ P
6 isinag.b ⊢ φ → B ∈ P
7 isinag.c ⊢ φ → C ∈ P
8 inagflat.g ⊢ φ → G ∈ 𝒢 Tarski
9 inagflat.x ⊢ φ → X ∈ P
10 inagflat.1 ⊢ φ → A ≠ B
11 inagflat.2 ⊢ φ → C ≠ B
12 inagflat.3 ⊢ φ → X ≠ B
13 inagflat.4 ⊢ φ → B ∈ A I C
14 eqidd ⊢ φ → B = B
15 14 orcd ⊢ φ → B = B ∨ B K ⁡ B X
16 1 2 3 4 5 6 7 8 6 10 11 12 13 15 isinagd ⊢ φ → X ∈ 𝒢 ∠ ⁡ G ⟨“ ABC ”⟩