Metamath Proof Explorer


Theorem hpgssplng

Description: Any point X on a half plane defined by a line A and another point Y is on the plane defined by A and Y . (Contributed by Thierry Arnoux, 5-Jul-2026)

Ref Expression
Hypotheses hpgssplng.p 𝑃 = ( Base ‘ 𝐺 )
hpgssplng.l 𝐿 = ( LineG ‘ 𝐺 )
hpgssplng.e 𝐸 = ( hlG ‘ 𝐺 )
hpgssplng.a ( 𝜑𝐴 ∈ ran 𝐿 )
hpgssplng.x ( 𝜑𝑋𝑃 )
hpgssplng.y ( 𝜑𝑌 ∈ ( 𝑃𝐴 ) )
hpgssplng.g ( 𝜑𝐺 ∈ TarskiG )
hpgssplng.1 ( 𝜑𝑋 ( ( hpG ‘ 𝐺 ) ‘ 𝐴 ) 𝑌 )
Assertion hpgssplng ( 𝜑𝑋 ∈ ( 𝐴 𝐸 𝑌 ) )

Proof

Step Hyp Ref Expression
1 hpgssplng.p 𝑃 = ( Base ‘ 𝐺 )
2 hpgssplng.l 𝐿 = ( LineG ‘ 𝐺 )
3 hpgssplng.e 𝐸 = ( hlG ‘ 𝐺 )
4 hpgssplng.a ( 𝜑𝐴 ∈ ran 𝐿 )
5 hpgssplng.x ( 𝜑𝑋𝑃 )
6 hpgssplng.y ( 𝜑𝑌 ∈ ( 𝑃𝐴 ) )
7 hpgssplng.g ( 𝜑𝐺 ∈ TarskiG )
8 hpgssplng.1 ( 𝜑𝑋 ( ( hpG ‘ 𝐺 ) ‘ 𝐴 ) 𝑌 )
9 8 3mix2d ( 𝜑 → ( 𝑋𝐴𝑋 ( ( hpG ‘ 𝐺 ) ‘ 𝐴 ) 𝑌𝑋 { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃𝐴 ) ∧ 𝑏 ∈ ( 𝑃𝐴 ) ) ∧ ∃ 𝑡𝐴 𝑡 ∈ ( 𝑎 ( Itv ‘ 𝐺 ) 𝑏 ) ) } 𝑌 ) )
10 eqid ( Itv ‘ 𝐺 ) = ( Itv ‘ 𝐺 )
11 eqid { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃𝐴 ) ∧ 𝑏 ∈ ( 𝑃𝐴 ) ) ∧ ∃ 𝑡𝐴 𝑡 ∈ ( 𝑎 ( Itv ‘ 𝐺 ) 𝑏 ) ) } = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃𝐴 ) ∧ 𝑏 ∈ ( 𝑃𝐴 ) ) ∧ ∃ 𝑡𝐴 𝑡 ∈ ( 𝑎 ( Itv ‘ 𝐺 ) 𝑏 ) ) }
12 1 10 2 3 7 4 6 11 5 elplng ( 𝜑 → ( 𝑋 ∈ ( 𝐴 𝐸 𝑌 ) ↔ ( 𝑋𝐴𝑋 ( ( hpG ‘ 𝐺 ) ‘ 𝐴 ) 𝑌𝑋 { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( ( 𝑎 ∈ ( 𝑃𝐴 ) ∧ 𝑏 ∈ ( 𝑃𝐴 ) ) ∧ ∃ 𝑡𝐴 𝑡 ∈ ( 𝑎 ( Itv ‘ 𝐺 ) 𝑏 ) ) } 𝑌 ) ) )
13 9 12 mpbird ( 𝜑𝑋 ∈ ( 𝐴 𝐸 𝑌 ) )