Metamath Proof Explorer


Theorem hpgssplng

Description: Any point X on a half plane defined by a line A and another point Y is on the plane defined by A and Y . (Contributed by Thierry Arnoux, 5-Jul-2026)

Ref Expression
Hypotheses hpgssplng.p ⊢ P = Base G
hpgssplng.l ⊢ L = Line 𝒢 ⁡ G
hpgssplng.e No typesetting found for |- E = ( PlnG ` G ) with typecode |-
hpgssplng.a ⊢ φ → A ∈ ran ⁡ L
hpgssplng.x ⊢ φ → X ∈ P
hpgssplng.y ⊢ φ → Y ∈ P ∖ A
hpgssplng.g ⊢ φ → G ∈ 𝒢 Tarski
hpgssplng.1 ⊢ φ → X hp 𝒢 ⁡ G ⁡ A Y
Assertion hpgssplng ⊢ φ → X ∈ A E Y

Proof

Step Hyp Ref Expression
1 hpgssplng.p ⊢ P = Base G
2 hpgssplng.l ⊢ L = Line 𝒢 ⁡ G
3 hpgssplng.e Could not format E = ( PlnG ` G ) : No typesetting found for |- E = ( PlnG ` G ) with typecode |-
4 hpgssplng.a ⊢ φ → A ∈ ran ⁡ L
5 hpgssplng.x ⊢ φ → X ∈ P
6 hpgssplng.y ⊢ φ → Y ∈ P ∖ A
7 hpgssplng.g ⊢ φ → G ∈ 𝒢 Tarski
8 hpgssplng.1 ⊢ φ → X hp 𝒢 ⁡ G ⁡ A Y
9 8 3mix2d ⊢ φ → X ∈ A ∨ X hp 𝒢 ⁡ G ⁡ A Y ∨ X a b | a ∈ P ∖ A ∧ b ∈ P ∖ A ∧ ∃ t ∈ A t ∈ a Itv ⁡ G b Y
10 eqid ⊢ Itv ⁡ G = Itv ⁡ G
11 eqid ⊢ a b | a ∈ P ∖ A ∧ b ∈ P ∖ A ∧ ∃ t ∈ A t ∈ a Itv ⁡ G b = a b | a ∈ P ∖ A ∧ b ∈ P ∖ A ∧ ∃ t ∈ A t ∈ a Itv ⁡ G b
12 1 10 2 3 7 4 6 11 5 elplng ⊢ φ → X ∈ A E Y ↔ X ∈ A ∨ X hp 𝒢 ⁡ G ⁡ A Y ∨ X a b | a ∈ P ∖ A ∧ b ∈ P ∖ A ∧ ∃ t ∈ A t ∈ a Itv ⁡ G b Y
13 9 12 mpbird ⊢ φ → X ∈ A E Y