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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 ) )