Looking at his book, and am obviously new to studying this. Could someone help explain to me how the truth table is valid here?
To my understanding, when $C=0$, the circuit behaves like a reversible AND gate, and when $C=1$, it acts as reversible NAND gate. So, my question is, shouldn't the output on the truth table be $f(1,1)$, when $A,B = 0$ and $C=1$?
Obviously, I'm wrong, but can't understand why.
The page for reference:
To add onto the answers, another way to look at why when A,B = 0 and C = 1, then $\overline{f(0,0)}$, is the inputs of f. The input of f is A and B. For example when A,B= 0, then f(0,0) will be in the result, because f takes in A and B as inputs(hence f(A,B) as denoted above). As others stated above, C = 1 then negates the problem, resulting in $\overline{f(0,0)}$. f(1,1) does not make sense because the inputs A,B = 0, not A,B = 1. If A,B = 1, THEN f(1,1) being in the result would make sense(in this case, the answer would be $\overline{f(1,1)}$, because C = 1 negates f). I hope this helps.
It should be $\overline{f(0,0)}$ when $A=B=0$ and $C=1$, I think. In the truth table, $f$ is unspecified but we could have $f$ be an AND gave and $\overline f$ be a NAND gate. But $f$ could be any of the two-input Boolean functions.
The inputs to $f$ are $A$ and $B$, while $C$ inverts whatever $f$ calculates.
If the function is $f(A, B)$, and $A = 0$ and $B = 0$, you just replace $A$ and $B$ for their corresponding values inside the function; therefore, $f(A, B)$ simply becomes $f(0, 0)$. Then, taking $f(0, 0) \oplus C$ for $C = 1$, turns $f(0, 0) \oplus 1$ into $\overline{f(0, 0})$ because adding a $1$ to a variable in modulo-2 is equivalent simply negating that variable.
Therefore, for $A = B = 0$ and $C = 1$:
$$f(A, B) \oplus C = \overline{f(0, 0})$$
