Metamath Proof Explorer


Theorem oppne1

Description: Points lying on opposite sides of a line cannot be on the line. (Contributed by Thierry Arnoux, 3-Mar-2020)

Ref Expression
Hypotheses hpg.p 𝑃 = ( Base ‘ 𝐺 )
hpg.d = ( dist ‘ 𝐺 )
hpg.i 𝐼 = ( Itv ‘ 𝐺 )
hpg.o 𝑂 = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃𝐷 ) ∧ 𝑏 ∈ ( 𝑃𝐷 ) ) ∧ ∃ 𝑡𝐷 𝑡 ∈ ( 𝑎 𝐼 𝑏 ) ) }
opphl.l 𝐿 = ( LineG ‘ 𝐺 )
opphl.d ( 𝜑𝐷 ∈ ran 𝐿 )
opphl.g ( 𝜑𝐺 ∈ TarskiG )
oppcom.a ( 𝜑𝐴𝑃 )
oppcom.b ( 𝜑𝐵𝑃 )
oppcom.o ( 𝜑𝐴 𝑂 𝐵 )
Assertion oppne1 ( 𝜑 → ¬ 𝐴𝐷 )

Proof

Step Hyp Ref Expression
1 hpg.p 𝑃 = ( Base ‘ 𝐺 )
2 hpg.d = ( dist ‘ 𝐺 )
3 hpg.i 𝐼 = ( Itv ‘ 𝐺 )
4 hpg.o 𝑂 = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃𝐷 ) ∧ 𝑏 ∈ ( 𝑃𝐷 ) ) ∧ ∃ 𝑡𝐷 𝑡 ∈ ( 𝑎 𝐼 𝑏 ) ) }
5 opphl.l 𝐿 = ( LineG ‘ 𝐺 )
6 opphl.d ( 𝜑𝐷 ∈ ran 𝐿 )
7 opphl.g ( 𝜑𝐺 ∈ TarskiG )
8 oppcom.a ( 𝜑𝐴𝑃 )
9 oppcom.b ( 𝜑𝐵𝑃 )
10 oppcom.o ( 𝜑𝐴 𝑂 𝐵 )
11 1 2 3 4 8 9 islnopp ( 𝜑 → ( 𝐴 𝑂 𝐵 ↔ ( ( ¬ 𝐴𝐷 ∧ ¬ 𝐵𝐷 ) ∧ ∃ 𝑡𝐷 𝑡 ∈ ( 𝐴 𝐼 𝐵 ) ) ) )
12 10 11 mpbid ( 𝜑 → ( ( ¬ 𝐴𝐷 ∧ ¬ 𝐵𝐷 ) ∧ ∃ 𝑡𝐷 𝑡 ∈ ( 𝐴 𝐼 𝐵 ) ) )
13 12 simplld ( 𝜑 → ¬ 𝐴𝐷 )