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 ( 𝜑𝐺 ∈ 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 ( 𝜑𝐺 ∈ TarskiG )
5 tglndim0.d ( 𝜑 → ( ♯ ‘ 𝐵 ) = 1 )
6 5 ad4antr ( ( ( ( ( 𝜑𝐴 ∈ ran 𝐿 ) ∧ 𝑥𝐵 ) ∧ 𝑦𝐵 ) ∧ ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) ) → ( ♯ ‘ 𝐵 ) = 1 )
7 simpllr ( ( ( ( ( 𝜑𝐴 ∈ ran 𝐿 ) ∧ 𝑥𝐵 ) ∧ 𝑦𝐵 ) ∧ ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) ) → 𝑥𝐵 )
8 simplr ( ( ( ( ( 𝜑𝐴 ∈ ran 𝐿 ) ∧ 𝑥𝐵 ) ∧ 𝑦𝐵 ) ∧ ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) ) → 𝑦𝐵 )
9 1 6 7 8 tgldim0eq ( ( ( ( ( 𝜑𝐴 ∈ ran 𝐿 ) ∧ 𝑥𝐵 ) ∧ 𝑦𝐵 ) ∧ ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) ) → 𝑥 = 𝑦 )
10 simprr ( ( ( ( ( 𝜑𝐴 ∈ ran 𝐿 ) ∧ 𝑥𝐵 ) ∧ 𝑦𝐵 ) ∧ ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) ) → 𝑥𝑦 )
11 9 10 pm2.21ddne ( ( ( ( ( 𝜑𝐴 ∈ ran 𝐿 ) ∧ 𝑥𝐵 ) ∧ 𝑦𝐵 ) ∧ ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) ) → ⊥ )
12 4 adantr ( ( 𝜑𝐴 ∈ ran 𝐿 ) → 𝐺 ∈ TarskiG )
13 simpr ( ( 𝜑𝐴 ∈ ran 𝐿 ) → 𝐴 ∈ ran 𝐿 )
14 1 2 3 12 13 tgisline ( ( 𝜑𝐴 ∈ ran 𝐿 ) → ∃ 𝑥𝐵𝑦𝐵 ( 𝐴 = ( 𝑥 𝐿 𝑦 ) ∧ 𝑥𝑦 ) )
15 11 14 r19.29vva ( ( 𝜑𝐴 ∈ ran 𝐿 ) → ⊥ )
16 15 inegd ( 𝜑 → ¬ 𝐴 ∈ ran 𝐿 )