A significant result in the Wilcoxon test rejects the null hypothesis that scores are distributed symmetrically around C. That leaves two alternatives: either they are distributed symmetrically around some other value, or they are distributed asymmetrically. If one starts with the assumption of symmetry, then rejection of the null hypothesis leaves only the possibility that scores are distributed symmetrically around some value other than C. But what can you conclude if you don't assume symmetry?
This question can be answered with the concept of the partial midrange. The midrange is defined as the point midway between the highest and lowest scores; if a set of scores ranges from 40 to 80 then its midrange is 60. We shall define one partial midrange as the point midway between the first and 99th percentile scores, define another partial midrange as the point midway between the second and 98th percentile scores, and so on. As the two points defining the partial midrange get closer and closer together, the partial midrange will approach the median, which is the 50th percentile score. We will consider the midrange and the median to be the extreme forms of a whole family of partial midranges.
Now suppose the Wilcoxon test gives a significant result in the positive direction, as for instance when 6 scores all fall above the hypothesized center C. With no assumption of symmetry, you can conclude that at least some of the partial midranges fall above C. With the assumption of symmetry, you could conclude that all the partial midranges fall above C. Thus you have either a vague conclusion ("at least some of the partial midranges") or a very specific conclusion ("all partial midranges"), depending on your assumption. It therefore follows that without the assumption of symmetry you can conclude that the center of the distribution exceeds C in some sense, while with the assumption you can conclude that the center exceeds C in every reasonable sense.
Another way to lessen the importance of symmetry in the Wilcoxon test is to define a parameter W as the median of all paired-case means. Consider for instance a population with only 3 cases, with scores of 3, 5, and 9. There are three possible pairs of cases: the 3-5 pair, the 3-9 pair, and the 5-9 pair. But we shall also allow each case to be paired with itself, so there is a 3-3 pair, a 5-5 pair, and a 9-9 pair. The means of these 6 pairs are respectively 4, 6, 7, 3, 5, 9. From low to high these means are 3, 4, 5, 6, 7, 9. W, the median of the paired-case means, is then 5.5. In all but the most unusual distributions, W falls between the mean and the median. In this 3-case example the mean is 5.67 and the median is 5, so this example follows the usual pattern.
Symmetry is not nearly so important in the Wilcoxon test if we think of it as a test on the population value of W. That is, the Wilcoxon test has good validity as a test on W, even if the distribution of scores is quite skewed. The Wilcoxon test is not a perfect test of W, but it is very good. Thus if you conclude for instance that W exceeds the hypothesized value C, it follows that in all probability either the mean or the median must exceed C, since one of those nearly always exceeds or at least equals W.