Metamath Proof Explorer


Theorem prlngpln

Description: Two parallel lines are on a common plane. (Contributed by Thierry Arnoux, 5-Jul-2026)

Ref Expression
Hypotheses prlngpln.l 𝐿 = ( LineG ‘ 𝐺 )
prlngpln.e 𝐸 = ( hlG ‘ 𝐺 )
prlngpln.p = ( parlnG ‘ 𝐺 )
prlngpln.g ( 𝜑𝐺𝑉 )
prlngpln.1 ( 𝜑𝐴 𝐵 )
prlngpln.2 ( 𝜑𝐴𝐵 )
Assertion prlngpln ( 𝜑 → ∃ ∈ ran 𝐸 ( 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 prlngpln.l 𝐿 = ( LineG ‘ 𝐺 )
2 prlngpln.e 𝐸 = ( hlG ‘ 𝐺 )
3 prlngpln.p = ( parlnG ‘ 𝐺 )
4 prlngpln.g ( 𝜑𝐺𝑉 )
5 prlngpln.1 ( 𝜑𝐴 𝐵 )
6 prlngpln.2 ( 𝜑𝐴𝐵 )
7 1 2 3 4 brprlng ( 𝜑 → ( 𝐴 𝐵 ↔ ( ( 𝐴 ∈ ran 𝐿𝐵 ∈ ran 𝐿 ) ∧ ( 𝐴 = 𝐵 ∨ ( ∃ ∈ ran 𝐸 ( 𝐴𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) ) ) ) )
8 5 7 mpbid ( 𝜑 → ( ( 𝐴 ∈ ran 𝐿𝐵 ∈ ran 𝐿 ) ∧ ( 𝐴 = 𝐵 ∨ ( ∃ ∈ ran 𝐸 ( 𝐴𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) ) ) )
9 8 simprd ( 𝜑 → ( 𝐴 = 𝐵 ∨ ( ∃ ∈ ran 𝐸 ( 𝐴𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) ) )
10 6 neneqd ( 𝜑 → ¬ 𝐴 = 𝐵 )
11 9 10 orcnd ( 𝜑 → ( ∃ ∈ ran 𝐸 ( 𝐴𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) )
12 11 simpld ( 𝜑 → ∃ ∈ ran 𝐸 ( 𝐴𝐵 ) )