The code can be simplified like this:

boolean condA = ( fooSuccess  && !barSuccess &&  mooSuccess )
boolean condB = ( fooSuccess  && !barSuccess && !mooSuccess )
boolean condC = (!fooSuccess  &&  barSuccess &&  mooSuccess )
boolean condD = (!fooSuccess  &&  barSuccess && !mooSuccess &&  cowSuccess)
boolean condE = (!fooSuccess  &&  barSuccess && !mooSuccess && !cowSuccess)

if (condA) {
    ....
    return;
}
if (condB) {
    ....
    return;
}

... and so son 

if (condE) {
    ....
    return;
}

In essence:

• Evaluate complex conditions into a single boolean
• Since conditions are exclusive (as it’s typically the case in nested arrow heads) you don’t have to nest the ifs, just add a return so that no other block is run.
• This reduces the cyclomatic complexity enormously and makes the code trivial.
• Having what exactly gets run in every case, you can even do a Karnaugh table to identify and remove redundant condition combinations, just as they do in logic circuits design. But you didn’t provide that in the example.
• Be sure to extract all this logic into its own method so the return statements don’t interfere with blocks that have to be run anymay.
• Descriptive names should be chosen for conA..condF.

user61852 has a pretty good answer solving the general problem of simplifying nested conditionals. However, I was intrigued with your proposed state machine-like solution, and wanted to illustrate how that can sometimes be preferable to a set of binary flags.

It all depends on the sequence of obtaining the flags’ values and how many combinations are valid. If you have n flags, you have 2ⁿ states. For 4 flags, that’s 16 states, and as you observed, only 3 of them may have any useful meaning. In situations like that, using a state variable instead can simplify things greatly.

Let’s say you have a lock that will open if 4 keys are pressed in the right order. If you press any key incorrectly, it immediately resets back to the start of the sequence. A very naïve way to implement a keypress handler using binary flags is:

void key_pressed(char key)
{
   if (!key1correct)
   {
      if (key == pin[0])
      {
         key1correct = true;
      }
   }
   else if (!key2correct)
   {
       if (key == pin[1])
       {
           key2correct = true;
       }
       else
       {
           key1correct = false;
       }
   }
   else if (!key3correct)
   {
       if (key == pin[2])
       {
           key3correct = true;
       }
       else
       {
           key1correct = false;
           key2correct = false;
       }
   }
   else
   {
       if (key == pin[3])
       {
           key4correct = true;
       }
       else
       {
           key1correct = false;
           key2correct = false;
           key3correct = false;
       }
   }

   if (key1correct && key2correct && key3correct && key4correct)
   {
       open_lock();
       key1correct = false;
       key2correct = false;
       key3correct = false;
       key4correct = false;
   }
}

A simplified, flattened version using binary flags (like user61852’s answer) looks like this:

void key_pressed(char key)
{
    if (!key1correct && key == pin[0])
    {
        key1correct = true;
        return;
    }

    if (key1correct && !key2correct && key == pin[1])
    {
        key2correct = true;
        return;
    }

    if (key1correct && key2correct && !key3correct && key == pin[2])
    {
        key3correct = true;
        return;
    }

    if (key1correct && key2correct && key3correct && key == pin[3])
    {
        open_lock();
        key1correct = false;
        key2correct = false;
        key3correct = false;
        return;
    }

    key1correct = false;
    key2correct = false;
    key3correct = false;
}

That’s a lot easier to read, but still in both of these solutions you have states like key2correct && !key1correct that are completely invalid and unused. Whereas using a state variable num_correct_keys instead of the binary flags looks like this:

void key_pressed(char key)
{
    if (key == pin[num_correct_keys])
        ++num_correct_keys;
    else
        num_correct_keys = 0;

    if (num_correct_keys == 4)
    {
        open_lock();
        num_correct_keys = 0;
    }
}

Granted, this is a contrived example, but people often write code like the first version in less obvious situations, sometimes spread across multiple files. Using a state variable often greatly simplifies code, especially when the flags represent a sequence of events.

I’m not sure your example is a particularly good one for this question, but it’s possible to condense it to be much simpler and still preserve the same logic:

if(fooSuccess && !barSuccess)
{
    if(mooSuccess)
    {
        // .....
    }
    else
    {
        // .....
    }
}
else if (!fooSuccess && barSuccess)
{
    if(mooSuccess)
    {
        // .....
    }
    else if(cowSuccess)
    {
        // .....
    }
    else 
    {
        // .....
    }
}

Notice how there’s no more than one level deep of ifs. To clean this up, I took the following steps:

• Never check whether a boolean value is == true, != true, or != false. Occasionally I’ve found it worth checking whether it == false to make it more apparent that that’s the expected case, but even that should be exceptional.
• If your else block consists of nothing but an if/else block, then you can merge it with the parent else.
• If your if statement only has one input/variable, you don’t need to check the inverse for the else.