for k, n, m being Nat holds
( (scf (m / n)) . k = (divSeq (m,n)) . k & (rfs (m / n)) . 1 = n / ((modSeq (m,n)) . 0) & (rfs (m / n)) . (k + 2) = ((modSeq (m,n)) . k) / ((modSeq (m,n)) . (k + 1)) )