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 = Line 𝒢 G
prlngpln3.e No typesetting found for |- E = ( PlnG ` G ) with typecode |-
prlngpln3.p No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
prlngpln3.g φ G 𝒢 Tarski
prlngpln3.1 φ A ˙ B
prlngpln3.2 φ A B
prlngpln3.x φ X B
Assertion prlngpln3 φ B A E X

Proof

Step Hyp Ref Expression
1 prlngpln3.l L = Line 𝒢 G
2 prlngpln3.e Could not format E = ( PlnG ` G ) : No typesetting found for |- E = ( PlnG ` G ) with typecode |-
3 prlngpln3.p Could not format .|| = ( parlnG ` G ) : No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
4 prlngpln3.g φ G 𝒢 Tarski
5 prlngpln3.1 φ A ˙ B
6 prlngpln3.2 φ A B
7 prlngpln3.x φ X B
8 eqid Base G = Base G
9 1 3 4 5 prlngrcl1 φ A ran L
10 9 adantr φ y B A ran L
11 eqid Itv G = Itv G
12 4 adantr φ y B G 𝒢 Tarski
13 5 adantr φ y B A ˙ B
14 1 3 12 13 prlngrcl2 φ y B B ran L
15 simpr φ y B y B
16 8 1 11 12 14 15 tglnpt φ y B y Base G
17 1 3 4 5 prlngrcl2 φ B ran L
18 8 1 11 4 17 7 tglnpt φ X Base G
19 incom A B = B A
20 1 3 4 5 6 prlngin0 φ A B =
21 19 20 eqtr3id φ B A =
22 disjel B A = X B ¬ X A
23 21 7 22 syl2anc φ ¬ X A
24 18 23 eldifd φ X Base G A
25 24 adantr φ y B X Base G A
26 6 adantr φ y B A B
27 7 adantr φ y B X B
28 1 2 3 12 13 26 15 27 prlnghpg φ y B y hp 𝒢 G A X
29 8 1 2 10 16 25 12 28 hpgssplng φ y B y A E X
30 29 ex φ y B y A E X
31 30 ssrdv φ B A E X