Metamath Proof Explorer


Theorem prlnginn0

Description: A line C intersecting another line A also intersects any line B parallel to A . Theorem 12.16 of Schwabhauser p. 125. (Contributed by Thierry Arnoux, 13-Jul-2026)

Ref Expression
Hypotheses prlnginn0.l ⊢ L = Line 𝒢 ⁡ G
prlnginn0.e No typesetting found for |- E = ( PlnG ` G ) with typecode |-
prlnginn0.p No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
prlnginn0.g ⊢ φ → G ∈ 𝒢 Tarski
prlnginn0.1 ⊢ φ → G ∈ 𝒢 Tarski E
prlnginn0.h ⊢ φ → H ∈ ran ⁡ E
prlnginn0.c ⊢ φ → C ∈ ran ⁡ L
prlnginn0.2 ⊢ φ → A ∩ C ≠ ∅
prlnginn0.3 ⊢ φ → A ≠ C
prlnginn0.4 ⊢ φ → A ∥ ˙ B
prlnginn0.5 ⊢ φ → A ⊆ H
prlnginn0.6 ⊢ φ → B ⊆ H
prlnginn0.7 ⊢ φ → C ⊆ H
Assertion prlnginn0 ⊢ φ → B ∩ C ≠ ∅

Proof

Step Hyp Ref Expression
1 prlnginn0.l ⊢ L = Line 𝒢 ⁡ G
2 prlnginn0.e Could not format E = ( PlnG ` G ) : No typesetting found for |- E = ( PlnG ` G ) with typecode |-
3 prlnginn0.p Could not format .|| = ( parlnG ` G ) : No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
4 prlnginn0.g ⊢ φ → G ∈ 𝒢 Tarski
5 prlnginn0.1 ⊢ φ → G ∈ 𝒢 Tarski E
6 prlnginn0.h ⊢ φ → H ∈ ran ⁡ E
7 prlnginn0.c ⊢ φ → C ∈ ran ⁡ L
8 prlnginn0.2 ⊢ φ → A ∩ C ≠ ∅
9 prlnginn0.3 ⊢ φ → A ≠ C
10 prlnginn0.4 ⊢ φ → A ∥ ˙ B
11 prlnginn0.5 ⊢ φ → A ⊆ H
12 prlnginn0.6 ⊢ φ → B ⊆ H
13 prlnginn0.7 ⊢ φ → C ⊆ H
14 8 neneqd ⊢ φ → ¬ A ∩ C = ∅
15 4 adantr ⊢ φ ∧ A ∥ ˙ C → G ∈ 𝒢 Tarski
16 simpr ⊢ φ ∧ A ∥ ˙ C → A ∥ ˙ C
17 9 adantr ⊢ φ ∧ A ∥ ˙ C → A ≠ C
18 1 3 15 16 17 prlngin0 ⊢ φ ∧ A ∥ ˙ C → A ∩ C = ∅
19 14 18 mtand ⊢ φ → ¬ A ∥ ˙ C
20 4 adantr ⊢ φ ∧ B ∥ ˙ C → G ∈ 𝒢 Tarski
21 6 adantr ⊢ φ ∧ B ∥ ˙ C → H ∈ ran ⁡ E
22 11 adantr ⊢ φ ∧ B ∥ ˙ C → A ⊆ H
23 10 adantr ⊢ φ ∧ B ∥ ˙ C → A ∥ ˙ B
24 5 adantr ⊢ φ ∧ B ∥ ˙ C → G ∈ 𝒢 Tarski E
25 13 adantr ⊢ φ ∧ B ∥ ˙ C → C ⊆ H
26 simpr ⊢ φ ∧ B ∥ ˙ C → B ∥ ˙ C
27 2 3 20 21 22 23 24 25 26 prlngplngtr ⊢ φ ∧ B ∥ ˙ C → A ∥ ˙ C
28 19 27 mtand ⊢ φ → ¬ B ∥ ˙ C
29 4 adantr ⊢ φ ∧ B ∩ C = ∅ → G ∈ 𝒢 Tarski
30 1 3 4 10 prlngrcl2 ⊢ φ → B ∈ ran ⁡ L
31 30 adantr ⊢ φ ∧ B ∩ C = ∅ → B ∈ ran ⁡ L
32 7 adantr ⊢ φ ∧ B ∩ C = ∅ → C ∈ ran ⁡ L
33 6 adantr ⊢ φ ∧ B ∩ C = ∅ → H ∈ ran ⁡ E
34 12 adantr ⊢ φ ∧ B ∩ C = ∅ → B ⊆ H
35 13 adantr ⊢ φ ∧ B ∩ C = ∅ → C ⊆ H
36 simpr ⊢ φ ∧ B ∩ C = ∅ → B ∩ C = ∅
37 1 2 3 29 31 32 33 34 35 36 prlngd ⊢ φ ∧ B ∩ C = ∅ → B ∥ ˙ C
38 28 37 mtand ⊢ φ → ¬ B ∩ C = ∅
39 38 neqned ⊢ φ → B ∩ C ≠ ∅