ABSTRACT
Proof. By definition 9, a * = ρ v N a * ∩ A ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0127.tif"/> . So if a* ≠ ∅ it must therefore either be the case since A ′ = ρ v N a ⊂ v N ∩ H ′ g H ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0128.tif"/> (equation 3.5) and ρ v N a * ⊂ ρ v N a ⊂ v N ⊂ H ′ R h h ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0129.tif"/> (definitions 5 and 6) that either ∃h ∈ a*, or ∃R hh′ ∈ a*. But similarly by that definition, if the latter is the case, R h h ′ ∈ ρ v N a * ⇒ h , h ′ ∈ ρ v N a * https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0130.tif"/> , hence we can conclude that if ∃ R h h ′ ∈ a * = ρ v N a * ∩ A ′ ⊂ ρ v N a * https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0131.tif"/> , then ∃ h ∈ ρ v N a * https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0132.tif"/> . Now since ρ v N a * ⊂ ρ v N a ⊂ v N https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0133.tif"/> by definition 9 it must be that h ∈ ρ v N a ⊂ v N https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0134.tif"/> , and by definition 6, ρ v N a ⊂ v N ⊂ ρ v N = H ′ R h h ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0135.tif"/> , so we must have h ∈ ρ (v N)\{R hh′} = H′. By equation 3.5, then h ∈ ρ v N a ⊂ v N https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0136.tif"/> and h ∈ H′, so h ∈ A′. Then we can conclude that since h ∈ ρ v N a * https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781351002301/825c3268-1059-4ac1-b4e2-58f88c33c172/content/ieq0137.tif"/> and h ∈ A′, we have h ∈ a*.
