theorem Th47:
  v9 = v & v1 <> v2 & (v = v1 or v = v2) & (the carrier' of G) in
  X implies Degree(v9, X) = Degree(v, X) +1
proof
  assume that
A1: v9 = v and
A2: v1 <> v2 and
A3: v = v1 or v = v2 and
A4: (the carrier' of G) in X;
  set E = the carrier' of G;
  per cases by A3;
  suppose
A5: v = v1;
    then
    Edges_In(v9, X) = Edges_In(v, X) & Edges_Out(v9, X) = Edges_Out(v, X)
    \/ {E} by A1,A2,A4,Th39,Th41;
    hence
    Degree(v9, X) = card Edges_In(v, X) + (card Edges_Out(v, X) + card {E
    }) by A1,A4,A5,Th41,CARD_2:40
      .= card Edges_In(v, X) + card Edges_Out(v, X) + card {E}
      .= Degree(v, X) +1 by CARD_1:30;
  end;
  suppose
A6: v = v2;
    then
    Edges_Out(v9, X) = Edges_Out(v, X) & Edges_In(v9, X) = Edges_In(v, X)
    \/ {E} by A1,A2,A4,Th40,Th42;
    hence
    Degree(v9, X) = card Edges_In(v, X) + card {E} + card Edges_Out(v, X)
    by A1,A4,A6,Th42,CARD_2:40
      .= card Edges_In(v, X) + card Edges_Out(v, X) + card {E}
      .= Degree(v, X) +1 by CARD_1:30;
  end;
end;
