Metamath Proof Explorer


Theorem nsnlplig

Description: There is no "one-point line" in a planar incidence geometry. (Contributed by BJ, 2-Dec-2021) (Proof shortened by AV, 5-Dec-2021)

Ref Expression
Assertion nsnlplig ( 𝐺 ∈ Plig → ¬ { 𝐴 } ∈ 𝐺 )

Proof

Step Hyp Ref Expression
1 eqid 𝐺 = 𝐺
2 1 l2p ( ( 𝐺 ∈ Plig ∧ { 𝐴 } ∈ 𝐺 ) → ∃ 𝑎 𝐺𝑏 𝐺 ( 𝑎𝑏𝑎 ∈ { 𝐴 } ∧ 𝑏 ∈ { 𝐴 } ) )
3 elsni ( 𝑎 ∈ { 𝐴 } → 𝑎 = 𝐴 )
4 elsni ( 𝑏 ∈ { 𝐴 } → 𝑏 = 𝐴 )
5 eqtr3 ( ( 𝑎 = 𝐴𝑏 = 𝐴 ) → 𝑎 = 𝑏 )
6 eqneqall ( 𝑎 = 𝑏 → ( 𝑎𝑏 → ¬ { 𝐴 } ∈ 𝐺 ) )
7 5 6 syl ( ( 𝑎 = 𝐴𝑏 = 𝐴 ) → ( 𝑎𝑏 → ¬ { 𝐴 } ∈ 𝐺 ) )
8 3 4 7 syl2an ( ( 𝑎 ∈ { 𝐴 } ∧ 𝑏 ∈ { 𝐴 } ) → ( 𝑎𝑏 → ¬ { 𝐴 } ∈ 𝐺 ) )
9 8 impcom ( ( 𝑎𝑏 ∧ ( 𝑎 ∈ { 𝐴 } ∧ 𝑏 ∈ { 𝐴 } ) ) → ¬ { 𝐴 } ∈ 𝐺 )
10 9 3impb ( ( 𝑎𝑏𝑎 ∈ { 𝐴 } ∧ 𝑏 ∈ { 𝐴 } ) → ¬ { 𝐴 } ∈ 𝐺 )
11 10 a1i ( ( 𝑎 𝐺𝑏 𝐺 ) → ( ( 𝑎𝑏𝑎 ∈ { 𝐴 } ∧ 𝑏 ∈ { 𝐴 } ) → ¬ { 𝐴 } ∈ 𝐺 ) )
12 11 rexlimivv ( ∃ 𝑎 𝐺𝑏 𝐺 ( 𝑎𝑏𝑎 ∈ { 𝐴 } ∧ 𝑏 ∈ { 𝐴 } ) → ¬ { 𝐴 } ∈ 𝐺 )
13 2 12 syl ( ( 𝐺 ∈ Plig ∧ { 𝐴 } ∈ 𝐺 ) → ¬ { 𝐴 } ∈ 𝐺 )
14 13 pm2.01da ( 𝐺 ∈ Plig → ¬ { 𝐴 } ∈ 𝐺 )