Metamath Proof Explorer


Theorem vtxdusgr0edgnel

Description: A vertex in a simple graph has degree 0 iff there is no edge incident with this vertex. (Contributed by AV, 17-Dec-2020) (Proof shortened by AV, 24-Dec-2020)

Ref Expression
Hypotheses vtxdushgrfvedg.v ⊢ 𝑉 = ( Vtx ‘ 𝐺 )
vtxdushgrfvedg.e ⊢ 𝐸 = ( Edg ‘ 𝐺 )
vtxdushgrfvedg.d ⊢ 𝐷 = ( VtxDeg ‘ 𝐺 )
Assertion vtxdusgr0edgnel ( ( 𝐺 ∈ USGraph ∧ 𝑈 ∈ 𝑉 ) → ( ( 𝐷 ‘ 𝑈 ) = 0 ↔ ¬ ∃ 𝑒 ∈ 𝐸 𝑈 ∈ 𝑒 ) )

Proof

Step Hyp Ref Expression
1 vtxdushgrfvedg.v ⊢ 𝑉 = ( Vtx ‘ 𝐺 )
2 vtxdushgrfvedg.e ⊢ 𝐸 = ( Edg ‘ 𝐺 )
3 vtxdushgrfvedg.d ⊢ 𝐷 = ( VtxDeg ‘ 𝐺 )
4 usgruhgr ⊢ ( 𝐺 ∈ USGraph → 𝐺 ∈ UHGraph )
5 1 2 3 vtxduhgr0edgnel ⊢ ( ( 𝐺 ∈ UHGraph ∧ 𝑈 ∈ 𝑉 ) → ( ( 𝐷 ‘ 𝑈 ) = 0 ↔ ¬ ∃ 𝑒 ∈ 𝐸 𝑈 ∈ 𝑒 ) )
6 4 5 sylan ⊢ ( ( 𝐺 ∈ USGraph ∧ 𝑈 ∈ 𝑉 ) → ( ( 𝐷 ‘ 𝑈 ) = 0 ↔ ¬ ∃ 𝑒 ∈ 𝐸 𝑈 ∈ 𝑒 ) )