Metamath Proof Explorer


Theorem usgr0edg0rusgr

Description: A simple graph is 0-regular iff it has no edges. (Contributed by Alexander van der Vekens, 12-Jul-2018) (Revised by AV, 19-Dec-2020) (Proof shortened by AV, 24-Dec-2020)

Ref Expression
Assertion usgr0edg0rusgr ( 𝐺 ∈ USGraph → ( 𝐺 RegUSGraph 0 ↔ ( Edg ‘ 𝐺 ) = ∅ ) )

Proof

Step Hyp Ref Expression
1 0nn0 0 ∈ ℕ0
2 isrusgr ( ( 𝐺 ∈ USGraph ∧ 0 ∈ ℕ0 ) → ( 𝐺 RegUSGraph 0 ↔ ( 𝐺 ∈ USGraph ∧ 𝐺 RegGraph 0 ) ) )
3 1 2 mpan2 ( 𝐺 ∈ USGraph → ( 𝐺 RegUSGraph 0 ↔ ( 𝐺 ∈ USGraph ∧ 𝐺 RegGraph 0 ) ) )
4 ibar ( 𝐺 ∈ USGraph → ( 𝐺 RegGraph 0 ↔ ( 𝐺 ∈ USGraph ∧ 𝐺 RegGraph 0 ) ) )
5 usgruhgr ( 𝐺 ∈ USGraph → 𝐺 ∈ UHGraph )
6 uhgr0edg0rgrb ( 𝐺 ∈ UHGraph → ( 𝐺 RegGraph 0 ↔ ( Edg ‘ 𝐺 ) = ∅ ) )
7 5 6 syl ( 𝐺 ∈ USGraph → ( 𝐺 RegGraph 0 ↔ ( Edg ‘ 𝐺 ) = ∅ ) )
8 3 4 7 3bitr2d ( 𝐺 ∈ USGraph → ( 𝐺 RegUSGraph 0 ↔ ( Edg ‘ 𝐺 ) = ∅ ) )