Metamath Proof Explorer


Theorem hpgne1

Description: Points on the open half plane cannot lie on its border. (Contributed by Thierry Arnoux, 4-Mar-2020)

Ref Expression
Hypotheses ishpg.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
ishpg.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
ishpg.l ⊢ 𝐿 = ( LineG ‘ 𝐺 )
ishpg.o ⊢ 𝑂 = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃 ∖ 𝐷 ) ∧ 𝑏 ∈ ( 𝑃 ∖ 𝐷 ) ) ∧ ∃ 𝑡 ∈ 𝐷 𝑡 ∈ ( 𝑎 𝐼 𝑏 ) ) }
ishpg.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
ishpg.d ⊢ ( 𝜑 → 𝐷 ∈ ran 𝐿 )
hpgbr.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
hpgbr.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
hpgne1.1 ⊢ ( 𝜑 → 𝐴 ( ( hpG ‘ 𝐺 ) ‘ 𝐷 ) 𝐵 )
Assertion hpgne1 ( 𝜑 → ¬ 𝐴 ∈ 𝐷 )

Proof

Step Hyp Ref Expression
1 ishpg.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
2 ishpg.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
3 ishpg.l ⊢ 𝐿 = ( LineG ‘ 𝐺 )
4 ishpg.o ⊢ 𝑂 = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃 ∖ 𝐷 ) ∧ 𝑏 ∈ ( 𝑃 ∖ 𝐷 ) ) ∧ ∃ 𝑡 ∈ 𝐷 𝑡 ∈ ( 𝑎 𝐼 𝑏 ) ) }
5 ishpg.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
6 ishpg.d ⊢ ( 𝜑 → 𝐷 ∈ ran 𝐿 )
7 hpgbr.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
8 hpgbr.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
9 hpgne1.1 ⊢ ( 𝜑 → 𝐴 ( ( hpG ‘ 𝐺 ) ‘ 𝐷 ) 𝐵 )
10 eqid ⊢ ( dist ‘ 𝐺 ) = ( dist ‘ 𝐺 )
11 6 ad2antrr ⊢ ( ( ( 𝜑 ∧ 𝑐 ∈ 𝑃 ) ∧ ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) → 𝐷 ∈ ran 𝐿 )
12 5 ad2antrr ⊢ ( ( ( 𝜑 ∧ 𝑐 ∈ 𝑃 ) ∧ ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) → 𝐺 ∈ TarskiG )
13 7 ad2antrr ⊢ ( ( ( 𝜑 ∧ 𝑐 ∈ 𝑃 ) ∧ ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) → 𝐴 ∈ 𝑃 )
14 simplr ⊢ ( ( ( 𝜑 ∧ 𝑐 ∈ 𝑃 ) ∧ ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) → 𝑐 ∈ 𝑃 )
15 simprl ⊢ ( ( ( 𝜑 ∧ 𝑐 ∈ 𝑃 ) ∧ ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) → 𝐴 𝑂 𝑐 )
16 1 10 2 4 3 11 12 13 14 15 oppne1 ⊢ ( ( ( 𝜑 ∧ 𝑐 ∈ 𝑃 ) ∧ ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) → ¬ 𝐴 ∈ 𝐷 )
17 1 2 3 4 5 6 7 8 hpgbr ⊢ ( 𝜑 → ( 𝐴 ( ( hpG ‘ 𝐺 ) ‘ 𝐷 ) 𝐵 ↔ ∃ 𝑐 ∈ 𝑃 ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) ) )
18 9 17 mpbid ⊢ ( 𝜑 → ∃ 𝑐 ∈ 𝑃 ( 𝐴 𝑂 𝑐 ∧ 𝐵 𝑂 𝑐 ) )
19 16 18 r19.29a ⊢ ( 𝜑 → ¬ 𝐴 ∈ 𝐷 )