The reason is that the subgraphs of G1 and G2 seen by Alice (resp., by Bob) may differ in more than one edge.
For example:

             Alice         Bob
            x     a      y     b
     G1 =   1 ... 0   |  0 ... 1
     G2 =   0 ... 1   |  1 ... 0
           ----------------------
     G  =   1 ... 1   |  1 ... 1

Here x and y are the free edges of a triangle in G. Now, if Alice "knows" that here edge x cannot appear together with the edge a,
and Bob "knows" that his edge y cannot appear together with the edge b, then they will (correctly) reject G.

To eliminate this "auxiliary knowledge" we must take

     G  =   1 ... 0   |  1 ... 0

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