‘Rooks console each other.’ (NRC Handelsblad, 23/01/2007) Question: can a rook console me? Can I console a rook? If in both cases the answer is ‘No’, then we have two isolated subframes, each characterised by a ‘console’-relation that has certain (formal and material) properties within that subframe, but which is limited to that subframe. The analogy with (in)translatability is obvious. The question that arises then is this: What reasons do we have (could we have) to call two such completely isolated relations both an ‘x-relation’? In what sense are these two the same relation?
Martin Stokhof from: Aantekeningen/Notes date: 23/01/2007
Part of the argumentation in Davidson’s ‘On the Very Idea of a Conceptual Scheme’ centers around transitivity of the translation relation. The point Davidson is trying to make here, I think, is that if there exists a translation from L1 into L2, that translation itself can be stated in either L1 or L2, and consists in an accurate mapping of sentences from L1 onto sentences of L2. If the translation is stated in L1 then, by assumption, it can also be stated in L2 (after all, there is a mapping from all of L1 into L2, i.e., also of the L1-sentences describing the mapping). So we can safely assume that if there is a translation from L1 into L2, it can be formulated in L2. Now suppose there is a translation from L2 into L3: that, again is a mapping, in this case of L2-sentences onto L3-sentences. That mapping includes those L2-sentences that describe how to translate L1-sentences into L2-sentences. So what we have then is (among other things) a set of L3-sentences that tell us which L2-sentences are translations of which L1-sentences, and L3-translations of the L2-sentences. But that means we have a translation of L1 into L3. (More neatly: if a translation is a homomorphism from L to L’, then we know that if there is a homomorphism from L1 to L2 and a homomorphism from L2 to L3, then there is a homomorphism from L1 to L3, viz.: the composition of the two.)
Of course the above argument only works if we assume that translations are total, i.e., that they map all sentences of L onto sentences of L’ and vice versa. What if we drop that assumption? First of all we have to ask ourselves whether we are still dealing with translation in such a case. But let that pass, and suppose we have a mapping of all the sentences of L1 onto a proper subset of sentences of L2, i.e., there are parts of L2 that have no counterpart in L1. Notice that the formulation of the translation can not be in the latter set (for that would mean that the translation would be statable in L2, but not in L1, which is absurd.) Now assume we have a translation from L2 into L3: the only way in which that would not give us a translation of L1 into L3 (by the reasoning above), is by being restricted to exactly that part of L2 that does not contain the L1-to-L2 translation. But that would mean that it really isn’t a translation of L2 as such: it must leave out a proper part of L2. Of course, such a mapping could exists, but we would lack any justification for calling it a translation. (And if we would insist that we could call it that, then we actually are assuming what we set out to prove, viz., the failure of transitivity of translation.)
Note that the fact that in actual cases there are bound to be discrepancies between languages, i.e., things in L that don’t have an exact counterpart in L’, or things in L’ that lack a counterpart in L, does not really affect this line of argumentation. The central point is that the translation itself concerns such a substantial part of both languages, that that by itself guarantees transitivity to a sufficient degree.
Martin Stokhof from: Radical Interpretation Discussion Board date: 11-2004