, Since |S| ? 4, this will actually leave at most two sums of S for the adjacent degree-2 vertices. We remove, in G, the vertices a 1 , a 2 , a 3 of any one initiating gadget, and, instead, identify its vertex e 1 and the vertex e 6 from a copy of the gadget H depicted in Figure 3. Now G has both vertices with degree 5 and 11, so the property above applies. Consider H. We claim that, in every neighbour-sum-distinguishing [2]-edge-weighting ? with sums from S (with |S | ? 3), neglecting e 6 , of H, necessarily 1) S must be {2, 3, 12}, 2) ?(e 5 e 6 ) = 1, and 3) ?(e 5 ) = 12. If this is true, note that, in G, a copy of H acts similarly as the three edges we have removed, G, without spoiling the existing degrees too much, new gadgets in such a way that G has, among others, vertices with degree 2, vertices with degree 5 and vertices with degree 11. Note that all these vertices are incompatible in terms of sums (when only weights 1 and 2 are used), in the sense that at least one sum in {2, vol.9

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