Transitions of Turing machine in Cook Levin theorem proof

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?

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: $$\{\overline{S[i, j, l]}, H[i, j], S[i+1, j, l]\}, 0\leqslant i
The group $$G_6$$ is made of the subgroup you quoted and the subgroup above.