Metamath Proof Explorer


Theorem isperp2d

Description: One direction of isperp2 . (Contributed by Thierry Arnoux, 10-Nov-2019)

Ref Expression
Hypotheses isperp.p ⊢ P = Base G
isperp.d ⊢ - ˙ = dist ⁡ G
isperp.i ⊢ I = Itv ⁡ G
isperp.l ⊢ L = Line 𝒢 ⁡ G
isperp.g ⊢ φ → G ∈ 𝒢 Tarski
isperp.a ⊢ φ → A ∈ ran ⁡ L
isperp2.b ⊢ φ → B ∈ ran ⁡ L
isperp2.x ⊢ φ → X ∈ A ∩ B
isperp2d.u ⊢ φ → U ∈ A
isperp2d.v ⊢ φ → V ∈ B
isperp2d.p ⊢ φ → A ⟂ 𝒢 ⁡ G B
Assertion isperp2d ⊢ φ → ⟨“ UXV ”⟩ ∈ ∟ 𝒢 ⁡ G

Proof

Step Hyp Ref Expression
1 isperp.p ⊢ P = Base G
2 isperp.d ⊢ - ˙ = dist ⁡ G
3 isperp.i ⊢ I = Itv ⁡ G
4 isperp.l ⊢ L = Line 𝒢 ⁡ G
5 isperp.g ⊢ φ → G ∈ 𝒢 Tarski
6 isperp.a ⊢ φ → A ∈ ran ⁡ L
7 isperp2.b ⊢ φ → B ∈ ran ⁡ L
8 isperp2.x ⊢ φ → X ∈ A ∩ B
9 isperp2d.u ⊢ φ → U ∈ A
10 isperp2d.v ⊢ φ → V ∈ B
11 isperp2d.p ⊢ φ → A ⟂ 𝒢 ⁡ G B
12 1 2 3 4 5 6 7 8 isperp2 ⊢ φ → A ⟂ 𝒢 ⁡ G B ↔ ∀ u ∈ A ∀ v ∈ B ⟨“ uXv ”⟩ ∈ ∟ 𝒢 ⁡ G
13 11 12 mpbid ⊢ φ → ∀ u ∈ A ∀ v ∈ B ⟨“ uXv ”⟩ ∈ ∟ 𝒢 ⁡ G
14 id ⊢ u = U → u = U
15 eqidd ⊢ u = U → X = X
16 eqidd ⊢ u = U → v = v
17 14 15 16 s3eqd ⊢ u = U → ⟨“ uXv ”⟩ = ⟨“ UXv ”⟩
18 17 eleq1d ⊢ u = U → ⟨“ uXv ”⟩ ∈ ∟ 𝒢 ⁡ G ↔ ⟨“ UXv ”⟩ ∈ ∟ 𝒢 ⁡ G
19 eqidd ⊢ v = V → U = U
20 eqidd ⊢ v = V → X = X
21 id ⊢ v = V → v = V
22 19 20 21 s3eqd ⊢ v = V → ⟨“ UXv ”⟩ = ⟨“ UXV ”⟩
23 22 eleq1d ⊢ v = V → ⟨“ UXv ”⟩ ∈ ∟ 𝒢 ⁡ G ↔ ⟨“ UXV ”⟩ ∈ ∟ 𝒢 ⁡ G
24 18 23 rspc2v ⊢ U ∈ A ∧ V ∈ B → ∀ u ∈ A ∀ v ∈ B ⟨“ uXv ”⟩ ∈ ∟ 𝒢 ⁡ G → ⟨“ UXV ”⟩ ∈ ∟ 𝒢 ⁡ G
25 9 10 24 syl2anc ⊢ φ → ∀ u ∈ A ∀ v ∈ B ⟨“ uXv ”⟩ ∈ ∟ 𝒢 ⁡ G → ⟨“ UXV ”⟩ ∈ ∟ 𝒢 ⁡ G
26 13 25 mpd ⊢ φ → ⟨“ UXV ”⟩ ∈ ∟ 𝒢 ⁡ G