Metamath Proof Explorer
Description: Deduce half-line from betweenness. (Contributed by Thierry Arnoux, 4-Mar-2020)
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|
Ref |
Expression |
|
Hypotheses |
ishlg.p |
⊢ 𝑃 = ( Base ‘ 𝐺 ) |
|
|
ishlg.i |
⊢ 𝐼 = ( Itv ‘ 𝐺 ) |
|
|
ishlg.k |
⊢ 𝐾 = ( hlG ‘ 𝐺 ) |
|
|
ishlg.a |
⊢ ( 𝜑 → 𝐴 ∈ 𝑃 ) |
|
|
ishlg.b |
⊢ ( 𝜑 → 𝐵 ∈ 𝑃 ) |
|
|
ishlg.c |
⊢ ( 𝜑 → 𝐶 ∈ 𝑃 ) |
|
|
hlln.1 |
⊢ ( 𝜑 → 𝐺 ∈ TarskiG ) |
|
|
hltr.d |
⊢ ( 𝜑 → 𝐷 ∈ 𝑃 ) |
|
|
btwnhl1.1 |
⊢ ( 𝜑 → 𝐶 ∈ ( 𝐴 𝐼 𝐵 ) ) |
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|
btwnhl1.2 |
⊢ ( 𝜑 → 𝐴 ≠ 𝐵 ) |
|
|
btwnhl1.3 |
⊢ ( 𝜑 → 𝐶 ≠ 𝐴 ) |
|
Assertion |
btwnhl1 |
⊢ ( 𝜑 → 𝐶 ( 𝐾 ‘ 𝐴 ) 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ishlg.p |
⊢ 𝑃 = ( Base ‘ 𝐺 ) |
2 |
|
ishlg.i |
⊢ 𝐼 = ( Itv ‘ 𝐺 ) |
3 |
|
ishlg.k |
⊢ 𝐾 = ( hlG ‘ 𝐺 ) |
4 |
|
ishlg.a |
⊢ ( 𝜑 → 𝐴 ∈ 𝑃 ) |
5 |
|
ishlg.b |
⊢ ( 𝜑 → 𝐵 ∈ 𝑃 ) |
6 |
|
ishlg.c |
⊢ ( 𝜑 → 𝐶 ∈ 𝑃 ) |
7 |
|
hlln.1 |
⊢ ( 𝜑 → 𝐺 ∈ TarskiG ) |
8 |
|
hltr.d |
⊢ ( 𝜑 → 𝐷 ∈ 𝑃 ) |
9 |
|
btwnhl1.1 |
⊢ ( 𝜑 → 𝐶 ∈ ( 𝐴 𝐼 𝐵 ) ) |
10 |
|
btwnhl1.2 |
⊢ ( 𝜑 → 𝐴 ≠ 𝐵 ) |
11 |
|
btwnhl1.3 |
⊢ ( 𝜑 → 𝐶 ≠ 𝐴 ) |
12 |
10
|
necomd |
⊢ ( 𝜑 → 𝐵 ≠ 𝐴 ) |
13 |
9
|
orcd |
⊢ ( 𝜑 → ( 𝐶 ∈ ( 𝐴 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐴 𝐼 𝐶 ) ) ) |
14 |
1 2 3 6 5 4 7
|
ishlg |
⊢ ( 𝜑 → ( 𝐶 ( 𝐾 ‘ 𝐴 ) 𝐵 ↔ ( 𝐶 ≠ 𝐴 ∧ 𝐵 ≠ 𝐴 ∧ ( 𝐶 ∈ ( 𝐴 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐴 𝐼 𝐶 ) ) ) ) ) |
15 |
11 12 13 14
|
mpbir3and |
⊢ ( 𝜑 → 𝐶 ( 𝐾 ‘ 𝐴 ) 𝐵 ) |