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Note carefully that we start the forward induction from χ∩ B(1) (i.e. the set of all core trajectory edges that have not been replaced by an exchange edge during the backward extension phase) and not B(1).
In conclusion, we add to our current set of edges all the compatible and exchange edges in H t and remove those (core) edges that are replaced by an exchange edge to arrive at B(t−1), formally: We apply the same procedure forward in time to obtain a sequence F(1)=χ∩ B(1), F(2),…, F(n) of forward extensions.
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Then, exchanging any edge e′ in the support of P with the edge e will produce a spanning tree for G V) of larger weight, implying that | e′|<| e| must hold for every such edge e′ implying that also B(P <| e| must hold.
Consequently, exchanging the edge e with the edge e i :={ v i −1, v i } in E would also give rise to a spanning tree for G V), and we would have | E′|=| E|+| e i |−| e|<| E| in view of | e i |≤ B(P <| e|, thus contradicting our choice of E. So, E⊆ E(V| W) must hold, as claimed.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com