Metamath Proof Explorer


Theorem tglnpt2

Description: Find a second point on a line. (Contributed by Thierry Arnoux, 18-Oct-2019)

Ref Expression
Hypotheses tglnpt2.p ⊢ P = Base G
tglnpt2.i ⊢ I = Itv ⁡ G
tglnpt2.l ⊢ L = Line 𝒢 ⁡ G
tglnpt2.g ⊢ φ → G ∈ 𝒢 Tarski
tglnpt2.a ⊢ φ → A ∈ ran ⁡ L
tglnpt2.x ⊢ φ → X ∈ A
Assertion tglnpt2 ⊢ φ → ∃ y ∈ A X ≠ y

Proof

Step Hyp Ref Expression
1 tglnpt2.p ⊢ P = Base G
2 tglnpt2.i ⊢ I = Itv ⁡ G
3 tglnpt2.l ⊢ L = Line 𝒢 ⁡ G
4 tglnpt2.g ⊢ φ → G ∈ 𝒢 Tarski
5 tglnpt2.a ⊢ φ → A ∈ ran ⁡ L
6 tglnpt2.x ⊢ φ → X ∈ A
7 neeq2 ⊢ y = z → X ≠ y ↔ X ≠ z
8 4 ad4antr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → G ∈ 𝒢 Tarski
9 simp-4r ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → x ∈ P
10 simpllr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → z ∈ P
11 simplrr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → x ≠ z
12 1 2 3 8 9 10 11 tglinerflx2 ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → z ∈ x L z
13 simplrl ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → A = x L z
14 12 13 eleqtrrd ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → z ∈ A
15 simpr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → X = x
16 15 11 eqnetrd ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → X ≠ z
17 7 14 16 rspcedvdw ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X = x → ∃ y ∈ A X ≠ y
18 neeq2 ⊢ y = x → X ≠ y ↔ X ≠ x
19 4 ad4antr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → G ∈ 𝒢 Tarski
20 simp-4r ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → x ∈ P
21 simpllr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → z ∈ P
22 simplrr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → x ≠ z
23 1 2 3 19 20 21 22 tglinerflx1 ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → x ∈ x L z
24 simplrl ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → A = x L z
25 23 24 eleqtrrd ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → x ∈ A
26 simpr ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → X ≠ x
27 18 25 26 rspcedvdw ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z ∧ X ≠ x → ∃ y ∈ A X ≠ y
28 17 27 pm2.61dane ⊢ φ ∧ x ∈ P ∧ z ∈ P ∧ A = x L z ∧ x ≠ z → ∃ y ∈ A X ≠ y
29 1 2 3 4 5 tgisline ⊢ φ → ∃ x ∈ P ∃ z ∈ P A = x L z ∧ x ≠ z
30 28 29 r19.29vva ⊢ φ → ∃ y ∈ A X ≠ y