Metamath Proof Explorer


Theorem prlngrcl1

Description: Reverse closure for parallelism. (Contributed by Thierry Arnoux, 5-Jul-2026)

Ref Expression
Hypotheses prlngin0.l 𝐿 = ( LineG ‘ 𝐺 )
prlngin0.p = ( parlnG ‘ 𝐺 )
prlngin0.g ( 𝜑𝐺𝑉 )
prlngin0.1 ( 𝜑𝐴 𝐵 )
Assertion prlngrcl1 ( 𝜑𝐴 ∈ ran 𝐿 )

Proof

Step Hyp Ref Expression
1 prlngin0.l 𝐿 = ( LineG ‘ 𝐺 )
2 prlngin0.p = ( parlnG ‘ 𝐺 )
3 prlngin0.g ( 𝜑𝐺𝑉 )
4 prlngin0.1 ( 𝜑𝐴 𝐵 )
5 eqid ( hlG ‘ 𝐺 ) = ( hlG ‘ 𝐺 )
6 1 5 2 3 brprlng ( 𝜑 → ( 𝐴 𝐵 ↔ ( ( 𝐴 ∈ ran 𝐿𝐵 ∈ ran 𝐿 ) ∧ ( 𝐴 = 𝐵 ∨ ( ∃ ∈ ran ( hlG ‘ 𝐺 ) ( 𝐴𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) ) ) ) )
7 4 6 mpbid ( 𝜑 → ( ( 𝐴 ∈ ran 𝐿𝐵 ∈ ran 𝐿 ) ∧ ( 𝐴 = 𝐵 ∨ ( ∃ ∈ ran ( hlG ‘ 𝐺 ) ( 𝐴𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) ) ) )
8 7 simplld ( 𝜑𝐴 ∈ ran 𝐿 )