Express state of overflow as function?

Question

Making an ALU I want a signal that signals whenever an addition or subtraction changes the binary number from - to +. Each number u and v is 4-bits and therefore is marked by 0, 1, 2, 3 where 3 is the most significant bit. c3 in this case is the carry in from the previous addition. z3 is the result and c4 is the potential carry out from the addition. I am using 2s complement, the list below explains how it works.

$$ 1000\hspace{1cm} -8\\ 1001\hspace{1cm} -7\\ 1010\hspace{1cm} -6\\ 1011\hspace{1cm} -5\\ 1100\hspace{1cm} -4\\ 1101\hspace{1cm} -3\\ 1110\hspace{1cm} -2\\ 1111\hspace{1cm} -1\\ 0000 \hspace{1.5cm} 0\\ 0001\hspace{1.5cm} 1\\ 0010\hspace{1.5cm} 2\\ 0011\hspace{1.5cm} 3\\ 0100\hspace{1.5cm} 4\\ 0101\hspace{1.5cm} 5\\ 0110\hspace{1.5cm} 6\\ 0111\hspace{1.5cm} 7\\ $$

As one can see here when adding for example 6 + 6. $0110ドル + 0110$$ The result is 1100 i.e -4. Since that isn't correct the signal OF (overflow) will turn one signaling this. As you can see below in the completed truth table, OF = 1 only when u3 and v3 are the same and the carry signals have alternate values. When adding 0110 + 0110 u3 and v3 is 0, c3 will be 1 since u2 and v2 will have a result of 0 and a carry out of 1.

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Now I wanna express the OF signal as a function. When trying I get this expression. $$OF = u_3 ́v_3 ́c_3c_4 ́z_3 \vee (u_3 ́v_3 ́c_3c_4 ́z_3) ́$$ This expression will always be equal to 1 since $$A \vee A ́ = 1$$ Therefore I'm a bit lost in how to express OF as a function.

Answer 1

Judging by your truth table for OF, it looks like you meant to write

$$OF = {u_3}' ,円 {v_3}' ,円 c_3 ,円 {c_4}' ,円 z_3 \vee u_3 ,円 v_3 ,円 {c_3}' ,円 c_4 ,円 {z_3}'$$

Remember that \$({u_3}' ,円 {v_3}' ,円 c_3 ,円 {c_4}' ,円 z_3)'\$ is not the same thing as \$u_3 ,円 v_3 ,円 {c_3}' ,円 c_4 ,円 {z_3}'\$.