Metamath Proof Explorer


Theorem perpln1

Description: Derive a line from perpendicularity. (Contributed by Thierry Arnoux, 27-Nov-2019)

Ref Expression
Hypotheses perpln.l ⊢ L = Line 𝒢 ⁡ G
perpln.1 ⊢ φ → G ∈ 𝒢 Tarski
perpln.2 ⊢ φ → A ⟂ 𝒢 ⁡ G B
Assertion perpln1 ⊢ φ → A ∈ ran ⁡ L

Proof

Step Hyp Ref Expression
1 perpln.l ⊢ L = Line 𝒢 ⁡ G
2 perpln.1 ⊢ φ → G ∈ 𝒢 Tarski
3 perpln.2 ⊢ φ → A ⟂ 𝒢 ⁡ G B
4 df-perpg ⊢ ⟂ 𝒢 = g ∈ V ⟼ a b | a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g
5 simpr ⊢ φ ∧ g = G → g = G
6 5 fveq2d ⊢ φ ∧ g = G → Line 𝒢 ⁡ g = Line 𝒢 ⁡ G
7 6 1 eqtr4di ⊢ φ ∧ g = G → Line 𝒢 ⁡ g = L
8 7 rneqd ⊢ φ ∧ g = G → ran ⁡ Line 𝒢 ⁡ g = ran ⁡ L
9 8 eleq2d ⊢ φ ∧ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g ↔ a ∈ ran ⁡ L
10 8 eleq2d ⊢ φ ∧ g = G → b ∈ ran ⁡ Line 𝒢 ⁡ g ↔ b ∈ ran ⁡ L
11 9 10 anbi12d ⊢ φ ∧ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ↔ a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L
12 5 fveq2d ⊢ φ ∧ g = G → ∟ 𝒢 ⁡ g = ∟ 𝒢 ⁡ G
13 12 eleq2d ⊢ φ ∧ g = G → ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
14 13 ralbidv ⊢ φ ∧ g = G → ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
15 14 rexralbidv ⊢ φ ∧ g = G → ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
16 11 15 anbi12d ⊢ φ ∧ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g ↔ a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
17 16 opabbidv ⊢ φ ∧ g = G → a b | a ∈ ran ⁡ Line 𝒢 ⁡ g ∧ b ∈ ran ⁡ Line 𝒢 ⁡ g ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ g = a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
18 2 elexd ⊢ φ → G ∈ V
19 1 fvexi ⊢ L ∈ V
20 rnexg ⊢ L ∈ V → ran ⁡ L ∈ V
21 19 20 mp1i ⊢ φ → ran ⁡ L ∈ V
22 21 21 xpexd ⊢ φ → ran ⁡ L × ran ⁡ L ∈ V
23 opabssxp ⊢ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ⊆ ran ⁡ L × ran ⁡ L
24 23 a1i ⊢ φ → a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ⊆ ran ⁡ L × ran ⁡ L
25 22 24 ssexd ⊢ φ → a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ∈ V
26 4 17 18 25 fvmptd2 ⊢ φ → ⟂ 𝒢 ⁡ G = a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
27 anass ⊢ a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ↔ a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
28 27 opabbii ⊢ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G = a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
29 26 28 eqtrdi ⊢ φ → ⟂ 𝒢 ⁡ G = a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
30 29 dmeqd ⊢ φ → dom ⁡ ⟂ 𝒢 ⁡ G = dom ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
31 dmopabss ⊢ dom ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G ⊆ ran ⁡ L
32 30 31 eqsstrdi ⊢ φ → dom ⁡ ⟂ 𝒢 ⁡ G ⊆ ran ⁡ L
33 relopabv ⊢ Rel ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
34 26 releqd ⊢ φ → Rel ⁡ ⟂ 𝒢 ⁡ G ↔ Rel ⁡ a b | a ∈ ran ⁡ L ∧ b ∈ ran ⁡ L ∧ ∃ x ∈ a ∩ b ∀ u ∈ a ∀ v ∈ b ⟨“ uxv ”⟩ ∈ ∟ 𝒢 ⁡ G
35 33 34 mpbiri ⊢ φ → Rel ⁡ ⟂ 𝒢 ⁡ G
36 brrelex12 ⊢ Rel ⁡ ⟂ 𝒢 ⁡ G ∧ A ⟂ 𝒢 ⁡ G B → A ∈ V ∧ B ∈ V
37 35 3 36 syl2anc ⊢ φ → A ∈ V ∧ B ∈ V
38 37 simpld ⊢ φ → A ∈ V
39 37 simprd ⊢ φ → B ∈ V
40 breldmg ⊢ A ∈ V ∧ B ∈ V ∧ A ⟂ 𝒢 ⁡ G B → A ∈ dom ⁡ ⟂ 𝒢 ⁡ G
41 38 39 3 40 syl3anc ⊢ φ → A ∈ dom ⁡ ⟂ 𝒢 ⁡ G
42 32 41 sseldd ⊢ φ → A ∈ ran ⁡ L