Transitions of Turing machine in Cook Levin theorem proof

Question

I am looking at the proof of the Cook-Levin theorem in Computers and Intractability: A Guide to the Theory of NP-Completeness. In particular, I find one thing unclear. He constructs the statement that is satisfied if and only if transition from step $i$ to step $i+1$ is correct. It goes as follows:

So essentially these statements say that when we are in state $q_k$ at time $i$ and the head is on the square $j$ and reads from it symbol $s_l$, then in the time $i+1$ our head will be at the square $j+\Delta$, we will be in state $k'$ and we will write $s_{l'}$ to the previous square. But I am worried since at no point we check what happens to the rest of the squares, namely if for example we had $ID_i$ to be $0000q_0000$ and $ID_{i+1}$ to be $11111q_511$, shouldn't all the conditions above be satisfied, however this isn't a valid transition?

Answer 1

This part is guaranteed by the first subgroup of $G_6$, described in the page 42:

The first subgroup guarantees that if the read-write head is not scanning tape square $j$ at time $i$, then the symbol in square $j$ does not change between times $i$ and $i+1$. The clauses in this subgroup are as follows: $$\{\overline{S[i, j, l]}, H[i, j], S[i+1, j, l]\}, 0\leqslant i<p(n), -p(n)\leqslant j\leqslant p(n) +1, 0\leqslant l\leqslant v$$

The group $G_6$ is made of the subgroup you quoted and the subgroup above.