Metamath Proof Explorer


Theorem prlngpln3

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

Ref Expression
Hypotheses prlngpln3.l 𝐿 = ( LineG ‘ 𝐺 )
prlngpln3.e 𝐸 = ( hlG ‘ 𝐺 )
prlngpln3.p = ( parlnG ‘ 𝐺 )
prlngpln3.g ( 𝜑𝐺 ∈ TarskiG )
prlngpln3.1 ( 𝜑𝐴 𝐵 )
prlngpln3.2 ( 𝜑𝐴𝐵 )
prlngpln3.x ( 𝜑𝑋𝐵 )
Assertion prlngpln3 ( 𝜑𝐵 ⊆ ( 𝐴 𝐸 𝑋 ) )

Proof

Step Hyp Ref Expression
1 prlngpln3.l 𝐿 = ( LineG ‘ 𝐺 )
2 prlngpln3.e 𝐸 = ( hlG ‘ 𝐺 )
3 prlngpln3.p = ( parlnG ‘ 𝐺 )
4 prlngpln3.g ( 𝜑𝐺 ∈ TarskiG )
5 prlngpln3.1 ( 𝜑𝐴 𝐵 )
6 prlngpln3.2 ( 𝜑𝐴𝐵 )
7 prlngpln3.x ( 𝜑𝑋𝐵 )
8 eqid ( Base ‘ 𝐺 ) = ( Base ‘ 𝐺 )
9 1 3 4 5 prlngrcl1 ( 𝜑𝐴 ∈ ran 𝐿 )
10 9 adantr ( ( 𝜑𝑦𝐵 ) → 𝐴 ∈ ran 𝐿 )
11 eqid ( Itv ‘ 𝐺 ) = ( Itv ‘ 𝐺 )
12 4 adantr ( ( 𝜑𝑦𝐵 ) → 𝐺 ∈ TarskiG )
13 5 adantr ( ( 𝜑𝑦𝐵 ) → 𝐴 𝐵 )
14 1 3 12 13 prlngrcl2 ( ( 𝜑𝑦𝐵 ) → 𝐵 ∈ ran 𝐿 )
15 simpr ( ( 𝜑𝑦𝐵 ) → 𝑦𝐵 )
16 8 1 11 12 14 15 tglnpt ( ( 𝜑𝑦𝐵 ) → 𝑦 ∈ ( Base ‘ 𝐺 ) )
17 1 3 4 5 prlngrcl2 ( 𝜑𝐵 ∈ ran 𝐿 )
18 8 1 11 4 17 7 tglnpt ( 𝜑𝑋 ∈ ( Base ‘ 𝐺 ) )
19 incom ( 𝐴𝐵 ) = ( 𝐵𝐴 )
20 1 3 4 5 6 prlngin0 ( 𝜑 → ( 𝐴𝐵 ) = ∅ )
21 19 20 eqtr3id ( 𝜑 → ( 𝐵𝐴 ) = ∅ )
22 disjel ( ( ( 𝐵𝐴 ) = ∅ ∧ 𝑋𝐵 ) → ¬ 𝑋𝐴 )
23 21 7 22 syl2anc ( 𝜑 → ¬ 𝑋𝐴 )
24 18 23 eldifd ( 𝜑𝑋 ∈ ( ( Base ‘ 𝐺 ) ∖ 𝐴 ) )
25 24 adantr ( ( 𝜑𝑦𝐵 ) → 𝑋 ∈ ( ( Base ‘ 𝐺 ) ∖ 𝐴 ) )
26 6 adantr ( ( 𝜑𝑦𝐵 ) → 𝐴𝐵 )
27 7 adantr ( ( 𝜑𝑦𝐵 ) → 𝑋𝐵 )
28 1 2 3 12 13 26 15 27 prlnghpg ( ( 𝜑𝑦𝐵 ) → 𝑦 ( ( hpG ‘ 𝐺 ) ‘ 𝐴 ) 𝑋 )
29 8 1 2 10 16 25 12 28 hpgssplng ( ( 𝜑𝑦𝐵 ) → 𝑦 ∈ ( 𝐴 𝐸 𝑋 ) )
30 29 ex ( 𝜑 → ( 𝑦𝐵𝑦 ∈ ( 𝐴 𝐸 𝑋 ) ) )
31 30 ssrdv ( 𝜑𝐵 ⊆ ( 𝐴 𝐸 𝑋 ) )