Metamath Proof Explorer


Theorem prlngpln4

Description: Building a parallel line conserves planes, i.e. given a line A and a point X not on A , the (unique) parallel B to A through X lies completely within the plane defined by A and X . Theorem 12.14 of Schwabhauser p. 124. (Contributed by Thierry Arnoux, 13-Jul-2026)

Ref Expression
Hypotheses prlngpln4.e No typesetting found for |- E = ( PlnG ` G ) with typecode |-
prlngpln4.p No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
prlngpln4.g ⊢ φ → G ∈ 𝒢 Tarski
prlngpln4.h ⊢ φ → H ∈ ran ⁡ E
prlngpln4.a ⊢ φ → A ⊆ H
prlngpln4.1 ⊢ φ → A ∥ ˙ B
prlngpln4.x ⊢ φ → X ∈ H
prlngpln4.2 ⊢ φ → X ∈ B
Assertion prlngpln4 ⊢ φ → B ⊆ H

Proof

Step Hyp Ref Expression
1 prlngpln4.e Could not format E = ( PlnG ` G ) : No typesetting found for |- E = ( PlnG ` G ) with typecode |-
2 prlngpln4.p Could not format .|| = ( parlnG ` G ) : No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
3 prlngpln4.g ⊢ φ → G ∈ 𝒢 Tarski
4 prlngpln4.h ⊢ φ → H ∈ ran ⁡ E
5 prlngpln4.a ⊢ φ → A ⊆ H
6 prlngpln4.1 ⊢ φ → A ∥ ˙ B
7 prlngpln4.x ⊢ φ → X ∈ H
8 prlngpln4.2 ⊢ φ → X ∈ B
9 simpr ⊢ φ ∧ A = B → A = B
10 5 adantr ⊢ φ ∧ A = B → A ⊆ H
11 9 10 eqsstrrd ⊢ φ ∧ A = B → B ⊆ H
12 eqid ⊢ Line 𝒢 ⁡ G = Line 𝒢 ⁡ G
13 3 adantr ⊢ φ ∧ A ≠ B → G ∈ 𝒢 Tarski
14 6 adantr ⊢ φ ∧ A ≠ B → A ∥ ˙ B
15 simpr ⊢ φ ∧ A ≠ B → A ≠ B
16 8 adantr ⊢ φ ∧ A ≠ B → X ∈ B
17 12 1 2 13 14 15 16 prlngpln3 ⊢ φ ∧ A ≠ B → B ⊆ A E X
18 eqid ⊢ Base G = Base G
19 4 adantr ⊢ φ ∧ A ≠ B → H ∈ ran ⁡ E
20 12 2 3 6 prlngrcl1 ⊢ φ → A ∈ ran ⁡ Line 𝒢 ⁡ G
21 20 adantr ⊢ φ ∧ A ≠ B → A ∈ ran ⁡ Line 𝒢 ⁡ G
22 7 adantr ⊢ φ ∧ A ≠ B → X ∈ H
23 12 2 13 14 15 prlngin0 ⊢ φ ∧ A ≠ B → A ∩ B = ∅
24 23 adantr ⊢ φ ∧ A ≠ B ∧ X ∈ A → A ∩ B = ∅
25 simpr ⊢ φ ∧ A ≠ B ∧ X ∈ A → X ∈ A
26 8 ad2antrr ⊢ φ ∧ A ≠ B ∧ X ∈ A → X ∈ B
27 25 26 elind ⊢ φ ∧ A ≠ B ∧ X ∈ A → X ∈ A ∩ B
28 27 ne0d ⊢ φ ∧ A ≠ B ∧ X ∈ A → A ∩ B ≠ ∅
29 28 neneqd ⊢ φ ∧ A ≠ B ∧ X ∈ A → ¬ A ∩ B = ∅
30 24 29 pm2.65da ⊢ φ ∧ A ≠ B → ¬ X ∈ A
31 22 30 eldifd ⊢ φ ∧ A ≠ B → X ∈ H ∖ A
32 5 adantr ⊢ φ ∧ A ≠ B → A ⊆ H
33 18 12 1 13 19 21 31 32 plng3p ⊢ φ ∧ A ≠ B → H = A E X
34 17 33 sseqtrrd ⊢ φ ∧ A ≠ B → B ⊆ H
35 11 34 pm2.61dane ⊢ φ → B ⊆ H