let IT1, IT2 be odd Element of NAT ; :: thesis: ( ( v in W .vertices() & IT1 <= len W & W . IT1 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT1 <= n ) & IT2 <= len W & W . IT2 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT2 <= n ) implies IT1 = IT2 ) & ( not v in W .vertices() & IT1 = len W & IT2 = len W implies IT1 = IT2 ) )

hereby :: thesis: ( not v in W .vertices() & IT1 = len W & IT2 = len W implies IT1 = IT2 )

assume v in W .vertices() ; :: thesis: ( IT1 <= len W & W . IT1 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT1 <= n ) & IT2 <= len W & W . IT2 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT2 <= n ) implies IT1 = IT2 )
assume A6: ( IT1 <= len W & W . IT1 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT1 <= n ) ) ; :: thesis: ( IT2 <= len W & W . IT2 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT2 <= n ) implies IT1 = IT2 )
assume A7: ( IT2 <= len W & W . IT2 = v & ( for n being odd Nat st n <= len W & W . n = v holds
IT2 <= n ) ) ; :: thesis: IT1 = IT2
then A8: IT1 <= IT2 by A6;
IT2 <= IT1 by A6, A7;
hence IT1 = IT2 by A8, XXREAL_0:1; :: thesis: verum

end;

assume not v in W .vertices() ; :: thesis: ( not IT1 = len W or not IT2 = len W or IT1 = IT2 )
thus ( not IT1 = len W or not IT2 = len W or IT1 = IT2 ) ; :: thesis: verum