Why does the ternary operator with commas evaluate only one expression in the true case?

I'm currently learning C++ with the book C++ Primer and one of the exercises in the book is:

Explain what the following expression does: someValue ? ++x, ++y : --x, --y

What do we know? We know that the ternary operator has a higher precedence than the comma operator. With binary operators this was quite easy to understand, but with the ternary operator I am struggling a bit. With binary operators "having higher precedence" means that we can use parentheses around the expression with higher precedence and it will not change the execution.

For the ternary operator I would do:

(someValue ? ++x, ++y : --x, --y)

effectively resulting in the same code which does not help me in understanding how the compiler will group the code.

However, from testing with a C++ compiler I know that the expression compiles and I do not know what a : operator could stand for by itself. So the compiler seems to interpret the ternary operator correctly.

Then I executed the program in two ways:

#include <iostream>

int main()
{
    bool someValue = true;
    int x = 10, y = 10;

    someValue ? ++x, ++y : --x, --y;

    std::cout << x << " " << y << std::endl;
    return 0;
}

Results in:

11 10

While on the other hand with someValue = false it prints:

9 9

Why would the C++ compiler generate code that for the true-branch of the ternary operator only increments x, while for the false-branch of the ternary it decrements both x and y?

I even went as far as putting parentheses around the true-branch like this:

someValue ? (++x, ++y) : --x, --y;

but it still results in 11 10.