 Solution

The shifts multiply the original array indices by four, and the following additions form byte memory addresses. The load into $s0 changes the variable $x$ itself, but the address of the next element of A has already been saved. The complete effect is

i = x
x = A[i]
B[y] = x + A[i + 1]

Here i is a fresh temporary. Both the assignment to $x$ and the assignment to $B[y]$ matter. Writing only $B[y]=A[x]+A[x+1]$ would omit the observable change to $x$. The interpretation assumes four-byte array elements, valid aligned addresses, and arithmetic within range; the source uses trapping signed additions. The address of $B[y]$ is computed before $x$ changes.