Metamath Proof Explorer


Theorem usgreqdrusgr

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* ∧ ∀ 𝑣𝑉 ( 𝐷𝑣 ) = 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* ∧ ∀ 𝑣𝑉 ( 𝐷𝑣 ) = 𝐾 ) ) )
4 3 3adant3 ( ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣𝑉 ( 𝐷𝑣 ) = 𝐾 ) → ( 𝐺 RegUSGraph 𝐾 ↔ ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣𝑉 ( 𝐷𝑣 ) = 𝐾 ) ) )
5 4 ibir ( ( 𝐺 ∈ USGraph ∧ 𝐾 ∈ ℕ0* ∧ ∀ 𝑣𝑉 ( 𝐷𝑣 ) = 𝐾 ) → 𝐺 RegUSGraph 𝐾 )