... is oddSo statement 1 follows immediately by conjugating statement 1 of Theorem 4.5. Whena is even, statement 6 follows by conjugating the formula in statement 6 of Theorem 4.5 and then substituting ... substitute n − k − 1 for k and k +1for n − k in statement 4 of Theorem 4.5 to obtain statement 5.Finally, we need to prove statement 5 of Lemma 4.7 and statements 6 and 7 of Theo-rem 4.5. Setting ... evens(1n)[s(2b−1,a+2)]ifa is oddSo statements 2 and 3 follow by substituting a for b − 1andb for a + 2 into statements 2 and 3 of Theorem 4.5.Finally, for statements 4 and 5, we haves(1k,n−k)[s(1a,b,b)]=s(1k,n−k)[s(2b−1,a+2)]ifa...