Let C be a c.r. code and (C,C1,C2,C3) the corresponding equitable partition with the intersection matrix Assume w.l.o.g. that 0000000... is a code word of C and 1000000..., 0100000..., 0010000..., 0001000..., 0000100..., 0000010... are its six neighbors from C1. Then the other k weight-1 words are in C. The word 1000000... has 5 neighbors from C2. These 5 words can be only 1100000..., 1010000..., 1001000..., 1000100..., 1000010..., as the other weight-2 words are at distance at most 1 from C. Similarly, all 15 weight-2 words with two ones in the first six positions are in C2.
Now consider 1000000...; it has two neighbors from C. One of them is 000..., the other has weight 2; assume w.l.o.g. that 10000010.... is in C.
Next consider 0100000...; it has two neighbors from C. One of them is 000..., the other has weight 2, 01000010.... or another, say w.l.o.g. 010000010.... Seeking a contradiction assume the last case holds. Since 10000010... and 010000010... are in C and 110000000... is in C2, the four words 100..., 010..., 11000010... and 110000010... belong to C1. So, 110000000... has 4 neighbors from C1; this contradicts the intersection array. So, 01000010... is in C.
Similarly, considering 0010000..., 0001000..., 0000100..., and 0000010..., we get that C contains 00100010..., 00010010..., 00001010..., and 00000110....
From this point, it is convenient to consider the c.r. code D obtained from C by inverting the 7th coordinate. We have shown that D contains 000..., 100..., 010..., 0010..., 00010..., 000010..., 0000010..., 00000010...; the 21 weight-2 words with ones in the first 7 positions belong to D1, and the other neighbors of 100..., 010..., 0010..., 00010..., 000010..., 0000010..., 00000010... belong to D. Similarly to the argument applied to C, we find that all weight-3 words with three ones in the first 7 positions belong to D2.
Each of them has two neighbors of weight 4 from D3. This means that the weight-4 words of D3 form a 3-(7,4,2) design in the first 7 positions (the weight-4 words with exactly 3 ones from {1,2,3,4,5,6,7} are at distance at most 2 from D, and so cannot belong to D3). Since there are no such design (a nontrivial t-(v,k,l) design must satisfy v-k>t), we find a CONTRADICTION.