Metamath Proof Explorer


Theorem prlngpln

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

Ref Expression
Hypotheses prlngpln.l ⊢ L = Line 𝒢 ⁡ G
prlngpln.e No typesetting found for |- E = ( PlnG ` G ) with typecode |-
prlngpln.p No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
prlngpln.g ⊢ φ → G ∈ V
prlngpln.1 ⊢ φ → A ∥ ˙ B
prlngpln.2 ⊢ φ → A ≠ B
Assertion prlngpln ⊢ φ → ∃ h ∈ ran ⁡ E A ⊆ h ∧ B ⊆ h

Proof

Step Hyp Ref Expression
1 prlngpln.l ⊢ L = Line 𝒢 ⁡ G
2 prlngpln.e Could not format E = ( PlnG ` G ) : No typesetting found for |- E = ( PlnG ` G ) with typecode |-
3 prlngpln.p Could not format .|| = ( parlnG ` G ) : No typesetting found for |- .|| = ( parlnG ` G ) with typecode |-
4 prlngpln.g ⊢ φ → G ∈ V
5 prlngpln.1 ⊢ φ → A ∥ ˙ B
6 prlngpln.2 ⊢ φ → A ≠ B
7 1 2 3 4 brprlng ⊢ φ → A ∥ ˙ B ↔ A ∈ ran ⁡ L ∧ B ∈ ran ⁡ L ∧ A = B ∨ ∃ h ∈ ran ⁡ E A ⊆ h ∧ B ⊆ h ∧ A ∩ B = ∅
8 5 7 mpbid ⊢ φ → A ∈ ran ⁡ L ∧ B ∈ ran ⁡ L ∧ A = B ∨ ∃ h ∈ ran ⁡ E A ⊆ h ∧ B ⊆ h ∧ A ∩ B = ∅
9 8 simprd ⊢ φ → A = B ∨ ∃ h ∈ ran ⁡ E A ⊆ h ∧ B ⊆ h ∧ A ∩ B = ∅
10 6 neneqd ⊢ φ → ¬ A = B
11 9 10 orcnd ⊢ φ → ∃ h ∈ ran ⁡ E A ⊆ h ∧ B ⊆ h ∧ A ∩ B = ∅
12 11 simpld ⊢ φ → ∃ h ∈ ran ⁡ E A ⊆ h ∧ B ⊆ h