Metamath Proof Explorer


Theorem tglndim0

Description: There are no lines in dimension 0. (Contributed by Thierry Arnoux, 18-Oct-2019)

Ref Expression
Hypotheses tglineelsb2.p ⊢ B = Base G
tglineelsb2.i ⊢ I = Itv ⁡ G
tglineelsb2.l ⊢ L = Line 𝒢 ⁡ G
tglineelsb2.g ⊢ φ → G ∈ 𝒢 Tarski
tglndim0.d ⊢ φ → B = 1
Assertion tglndim0 ⊢ φ → ¬ A ∈ ran ⁡ L

Proof

Step Hyp Ref Expression
1 tglineelsb2.p ⊢ B = Base G
2 tglineelsb2.i ⊢ I = Itv ⁡ G
3 tglineelsb2.l ⊢ L = Line 𝒢 ⁡ G
4 tglineelsb2.g ⊢ φ → G ∈ 𝒢 Tarski
5 tglndim0.d ⊢ φ → B = 1
6 5 ad4antr ⊢ φ ∧ A ∈ ran ⁡ L ∧ x ∈ B ∧ y ∈ B ∧ A = x L y ∧ x ≠ y → B = 1
7 simpllr ⊢ φ ∧ A ∈ ran ⁡ L ∧ x ∈ B ∧ y ∈ B ∧ A = x L y ∧ x ≠ y → x ∈ B
8 simplr ⊢ φ ∧ A ∈ ran ⁡ L ∧ x ∈ B ∧ y ∈ B ∧ A = x L y ∧ x ≠ y → y ∈ B
9 1 6 7 8 tgldim0eq ⊢ φ ∧ A ∈ ran ⁡ L ∧ x ∈ B ∧ y ∈ B ∧ A = x L y ∧ x ≠ y → x = y
10 simprr ⊢ φ ∧ A ∈ ran ⁡ L ∧ x ∈ B ∧ y ∈ B ∧ A = x L y ∧ x ≠ y → x ≠ y
11 9 10 pm2.21ddne ⊢ φ ∧ A ∈ ran ⁡ L ∧ x ∈ B ∧ y ∈ B ∧ A = x L y ∧ x ≠ y → ⊥
12 4 adantr ⊢ φ ∧ A ∈ ran ⁡ L → G ∈ 𝒢 Tarski
13 simpr ⊢ φ ∧ A ∈ ran ⁡ L → A ∈ ran ⁡ L
14 1 2 3 12 13 tgisline ⊢ φ ∧ A ∈ ran ⁡ L → ∃ x ∈ B ∃ y ∈ B A = x L y ∧ x ≠ y
15 11 14 r19.29vva ⊢ φ ∧ A ∈ ran ⁡ L → ⊥
16 15 inegd ⊢ φ → ¬ A ∈ ran ⁡ L