let IT1, IT2 be odd Element of NAT ; :: thesis: ( ( v in W .vertices() & IT1 <= len W & W . IT1 = v & ( for n being odd Element of NAT st n <= len W & W . n = v holds
n <= IT1 ) & IT2 <= len W & W . IT2 = v & ( for n being odd Element of NAT st n <= len W & W . n = v holds
n <= IT2 ) 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 Element of NAT st n <= len W & W . n = v holds
n <= IT1 ) & IT2 <= len W & W . IT2 = v & ( for n being odd Element of NAT st n <= len W & W . n = v holds
n <= IT2 ) implies IT1 = IT2 )
assume A5: ( IT1 <= len W & W . IT1 = v & ( for n being odd Element of NAT st n <= len W & W . n = v holds
n <= IT1 ) ) ; :: thesis: ( IT2 <= len W & W . IT2 = v & ( for n being odd Element of NAT st n <= len W & W . n = v holds
n <= IT2 ) implies IT1 = IT2 )
assume A6: ( IT2 <= len W & W . IT2 = v & ( for n being odd Element of NAT st n <= len W & W . n = v holds
n <= IT2 ) ) ; :: thesis: IT1 = IT2
then A7: IT2 <= IT1 by A5;
IT1 <= IT2 by A5, A6;
hence IT1 = IT2 by A7, XXREAL_0:1; :: thesis: verum

end;

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