Fuzzy logic and Truth Values Let's say we have the three-valued fuzzy logic with truth-values from $Q =$ {$0, 1/2, 1$}. My question is can the set of {$∧, ∨, ¬, ⊕, ↔,→$} plus the three truth-values be used to generate all functions from $Q^2 → Q$? 
Operators are defined: 
$p ∧ q = min${$p, q$};
$p ∨ q = max${$p, q$};
$¬p = 1 − p$;
$p → q = min${$1 + q − p, 1$};
$p ↔ q = min${$1 + p − q, 1 + q − p$};
$p ⊕ q = min${$p + q, 2 − p − q$}.
 A: Without using the truth-values as constants, you can't capture any function that returns the value of $\frac{1}{2}$ for one or more of the cases where all arguments have a value of $0$ or $1$, since no matter which and how many of the operators you use, the result will always be $0$ or $1$
But if you have the truth-functions as constants, then yes, you can capture any function (with any number of variables) with these. In fact, all you need is the constant $\frac{1}{2}$
To see why, let's use the same strategy as we do with boolean logic: take any truth-function, put it on a truth-table, and generate terms for each row that in the end we all disjunct together. This time, though, we need to capture the rows where it has a value of $1$ as well as the rows where it has a truth-value of $\frac{1}{2}$.
Now, to pick out a specific row, we want to able to say things like 'the row where $p=\frac{1}{2}$ and $q = 1$'. OK, so what we really want is an expression that returns $1$ where $p=\frac{1}{2}$ and returns $0$ otherwise, and same for $p=1$ and $p=0$
Well, for $p=\frac{1}{2}$ we can use $p \leftrightarrow \neg p$, for $p=1$ we can use $\neg (p \rightarrow \neg p)$, and for $p =0$ we can use $\neg (\neg p \rightarrow \neg p)$
OK, so with that, and using the conjunction to do this for all variables, we can pick out exactly the row we want, i.e. we get an expression that returns a $1$ for that row, and no other rows. Then, when the truth-function for that row is $\frac{1}{2}$, we simply conjunct that expression with $\frac{1}{2}$. And finally, we disjunct all the resulting expressions together.
Example. Suppose we want to capture the following function:
\begin{array}{cc|c}
p&q&f(p,q)\\
\hline
1&1&0\\
1&\frac{1}{2}&0\\
1&0&1\\
\frac{1}{2}&1&\frac{1}{2}\\
\frac{1}{2}&\frac{1}{2}&0\\
\frac{1}{2}&0&1\\
0&1&0\\
0&\frac{1}{2}&0\\
0&0&\frac{1}{2}\\
\end{array}
Then using the method as described above we get the expression:
$$(\neg (p \rightarrow \neg p) \land \neg (\neg q \rightarrow q)) \lor $$
$$((p \leftrightarrow \neg p) \land \neg (q \rightarrow \neg q) \land \frac{1}{2}) \lor $$
$$((p \leftrightarrow \neg p) \land \neg (\neg q \rightarrow q)) \lor$$
$$(\neg (\neg p \rightarrow p) \land \neg (\neg q \rightarrow q) \land \frac{1}{2})$$
