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 = ( LineG ‘ 𝐺 )
tglineelsb2.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
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 = ( LineG ‘ 𝐺 )
4 tglineelsb2.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
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 ∈ TarskiG )
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 )