Metamath Proof Explorer
Description: If all vertices in a simple graph have the same degree, the graph is
k-regular. (Contributed by AV, 26-Dec-2020)
|
|
Ref |
Expression |
|
Hypotheses |
isrusgr0.v |
⊢ 𝑉 = ( Vtx ‘ 𝐺 ) |
|
|
isrusgr0.d |
⊢ 𝐷 = ( VtxDeg ‘ 𝐺 ) |
|
Assertion |
usgreqdrusgr |
⊢ ( ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣 ∈ 𝑉 ( 𝐷 ‘ 𝑣 ) = 𝐾 ) → 𝐺 RegUSGraph 𝐾 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
isrusgr0.v |
⊢ 𝑉 = ( Vtx ‘ 𝐺 ) |
2 |
|
isrusgr0.d |
⊢ 𝐷 = ( VtxDeg ‘ 𝐺 ) |
3 |
1 2
|
isrusgr0 |
⊢ ( ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ) → ( 𝐺 RegUSGraph 𝐾 ↔ ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣 ∈ 𝑉 ( 𝐷 ‘ 𝑣 ) = 𝐾 ) ) ) |
4 |
3
|
3adant3 |
⊢ ( ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣 ∈ 𝑉 ( 𝐷 ‘ 𝑣 ) = 𝐾 ) → ( 𝐺 RegUSGraph 𝐾 ↔ ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣 ∈ 𝑉 ( 𝐷 ‘ 𝑣 ) = 𝐾 ) ) ) |
5 |
4
|
ibir |
⊢ ( ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣 ∈ 𝑉 ( 𝐷 ‘ 𝑣 ) = 𝐾 ) → 𝐺 RegUSGraph 𝐾 ) |