indicated statement will not run

I have run into an issue where my program for some reason cannot run an if statement, my code is posted below and the statement that will not run is 'if o == 2;' (arrowed to) The code is up to over 2000 permeatations so I can't give the array in use. If anyone can tell why it would be appreciated.

a = 1;
b = 1;
c = 1;
d = 1;
e = 1;
f = 1;

v = 1;
w = 1;
x = 1;

o = 0;

if true  
    if b <= length(B);
        while b <= length(B);
            V{v} = A{a} + B{b};
            b = b + 1;
            v = v + 1;
        end
    end
    while true
        if b > length(B);
            b = 1;
            a = a + 1;
            if a <= length(A);
                while b <= length(B);
                    V{v} = A{a} + B{b};
                    b = b + 1;
                    v = v + 1;
                end
            else 
                o = o + 1;
            end
        end
        if o == 1;
            if d <= length(D);
                while d <= length(D);
                    W{w} = C{c} + D{d};
                    d = d + 1;
                    w = w + 1;
                end
            end 
    
            while true
                if d > length(D);
                    d = 1;
                    c = c + 1;
                    if c <= length(C);
                        while d <= length(D);
                            W{w} = C{c} + D{d};
                            d = d + 1;
                            w = w + 1;
                        end
                    else 
                        o = o + 1
                    end
                end
            end
        end
        if o == 2;      %<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
            if f <= length(F)
                while f <= length(F)
                    X{x} = E{e} + F{f};
                    f = f + 1;
                    x = x + 1;
                end
            end
            while true
                if f > length(F)
                    f = 1;
                    e = e + 1;
                    if e <= length(E)
                        while f <= length(F)
                            X{x} = E{e} + F{f};
                            f = f + 1;
                            x = x + 1;
                        end
                    else
                        o = o + 1;
                        return
                    end
                end
            end
    
        end
    end
end

Answer 1

[Walter Roberson]

        if o == 1;
            if d <= length(D);
                while d <= length(D);
                    W{w} = C{c} + D{d};
                    d = d + 1;
                    w = w + 1;
                end
            end 

When you get out of that, d > length(D), either because d <= length(D) was originally false, or because the while loop incremented d until it became > length(D)

            while true
                if d > length(D);

The first time entering that loop, d > length(D) for the reasons described above

                    d = 1;
                    c = c + 1;

Suppose you just incremented c to be > length(C )

                    if c <= length(C);

so that is false

                        while d <= length(D);

so that is not done

                            W{w} = C{c} + D{d};
                            d = d + 1;
                            w = w + 1;
                        end
                    else

the else branch is taken because c > length(C)

 
                        o = o + 1

so o is incremented

                    end
                end
            end

You are inside a while true so you do the loop again. You assigned 1 to d the last time through, so d > length(D) is false, so you fall through to the bottom of the if and the end of the while true is immediately after that so nothing is done before starting the while true again. But d > length(D) is still false, so you do nothing before starting the while true again...