You can't directly compare an int[] to an int. You need to access a value in the array and then compare.

int[] myArray = {1, 2, 3};
int value = 1;
if (myArray[0] == value) {/*do something*/}

First things first, you need to loop to < a.length rather than a.length - 1. As this is strictly less than you need to include the upper bound.

So, to check all pairs of elements you can do:

for (int i = 0; i < a.length; i++) {
    for (int k = 0; k < a.length; k++) {
        if (a[i] != a[k]) {
            //do stuff
        }
    }
}

But this will compare, for example a[2] to a[3] and then a[3] to a[2]. Given that you are checking != this seems wasteful.

A better approach would be to compare each element i to the rest of the array:

for (int i = 0; i < a.length; i++) {
    for (int k = i + 1; k < a.length; k++) {
        if (a[i] != a[k]) {
            //do stuff
        }
    }
}

So if you have the indices [1...5] the comparison would go

1. 1 -> 2
2. 1 -> 3
3. 1 -> 4
4. 1 -> 5
5. 2 -> 3
6. 2 -> 4
7. 2 -> 5
8. 3 -> 4
9. 3 -> 5
10. 4 -> 5

So you see pairs aren't repeated. Think of a circle of people all needing to shake hands with each other.

If it's really an array, you want:

if (list[0] == list[2] && list[1] == list[3])

Note that if the array is of reference types, that's comparing by reference identity rather than for equality. You might want:

if (list[0].equals(list[2])) && list[1].equals(list[3]))

Although that will then go bang if any of the values is null. You might want a helper method to cope with this:

public static objectsEqual(Object o1, Object o2)
{
    if (o1 == o2)
    {
        return true;
    }
    if (o1 == null || o2 == null)
    {
        return false;
    }
    return o1.equals(o2);
}

Then:

if (objectsEqual(list[0], list[2]) && objectsEqual(list[1], list[3]))

If you've really got an ArrayList instead of an array then all of the above still holds, just using list.get(x) instead of list[x] in each place.