Mathematical logic - Adequate sets of connectives.

Question

How can I prove that $$\{\sim,\leftrightarrow \}$$ is not an adequate set of connectives?

Please help me with this exercise, I cannot prove it.

Answer 1

If you already know that $\{\lnot, \oplus\}$ is not adequate where $\oplus$ is xor, then you could note that $$ x \leftrightarrow y \text{ is equivalent to } x \oplus \lnot y. $$ Therefore, any expression using $\{\lnot, \leftrightarrow\}$ can also be expressed with $\{\lnot, \oplus\}$, and since the latter can't express everything, the former can't express everything either.

If you don't know that, maybe you have to come up with a proof from scratch that $\{\lnot, \leftrightarrow\}$ can't express everything. To do so, we have to think of some property that expressions using $\{\lnot, \leftrightarrow\}$ have -- some restrictive property that makes these expressions special. How about:

• In any expression using $\lnot$ and $\leftrightarrow$ only, for any variable $x$ in the expression, switching the value of a variable $x$ (from true to false or vice versa) EITHER (i) always keeps the truth value of the expression the same, OR (ii) always switches the truth value of the expression.

Can you show that this property is true?

Now we have our special restriction, and this will help us prove $\{\lnot, \leftrightarrow\}$ is not adequate. Assume, towards contradiction, that they are adequate. Then they can express $x \to y$ (if $x$ then $y$). So some expression of $x$ and $y$, using just $\{\lnot, \leftrightarrow\}$, is equivalent to $x \to y$.

Plug in T for $x$ and $y$, and we get $T \to T = T$. Switch the value of $x$ to F and we must get $F \to T = T$ -- so, our special property tells us that switching the value of $x$ always keeps the truth value the same. On the other hand, switching the value of $y$ to $F$, we get $T \to F = F$. So our special property tells us that switching the value of $y$ always switches the truth value of the expression.

But what happens if we switch both $x$ and $y$ to $F$? Then we get $F \to F = T$. But we switched $x$ once and $y$ once, so the expression should switch once and stay the same once, and it started out as $T$. So it should end up at $F$. This is a contradiction.

Thus we have proven that $\{\lnot, \leftrightarrow\}$ is not adequate.

Answer 2

Claim: Any truth-function $\phi$ defined over two or more variables and using $\neg$ and $\leftrightarrow$ only will have an even number of $T$'s and (therefore) an even number of $F$'s in the truth-table for $\phi$.

Proof: Take any such truth-function $\phi$. Since it involves two or more variables, the number of rows in the truth-table is a multiple of 4. By Induction over structure of $\phi$ we'll show that any subformula of $\phi$ will have an even number of $T$'s and $F$'s in the truth-table of $\phi$

Base: Take atomic statement $P$. In the truth-table of $\phi$, exactly half of the times $P$ will be $T$, and the other half it is $F$ (remember: we're looking at what is under the $P$ column in the truth-table for $\phi$). So given that the number of rows in the truth-table is a multiple of 4, there are an even number of $T$'s and an even number of $F$'s

Step: Let $\psi$ be a subformula of $\phi$. We need to consider two cases:

Case 1: $\psi = \neg \psi_1$

By inductive hypothesis, $\psi_1$ has an even number of $T$'s and $F$'s in the truth-table. Since all $T$'s become $F$'s and vice versa when negating, that means that $\psi$ has an even number of $T$'s and $F$'s in the truth-table.

Case 2: $\psi = \psi_1 \leftrightarrow \psi_2$

By inductive hypothesis, $\psi_1$ and $\psi_2$ both have an even number of $T$'s and $F$'s in the truth-table.

Now consider what happens when we evaluate $\psi = \psi_1 \leftrightarrow \psi_2$. Let us first consider the $m$ rows where $\psi_1$ is $T$. Of those rows, assume that $\psi_2$ is $T$ in $m_1$ of those and hence $F$ in $m-m_1$ of those. This gives us $m_1$ $T$'s and $m-m_1$ $F$'s for $\psi$. Now consider the $n$ rows where $\psi_1$ is $F$. Of those rows, assume that $\psi_2$ is $T$ in $n_1$ of those and hence $F$ in $n-n_1$ of those. This gives us $n_1$ $F$'s and $n-n_1$ $T$'s for $\psi$. So, in total we get $m_1 + p_2$ $T$'s and $m_2 + p_1$ $F$'s for $\psi$.

But, since by inductive hypothesis $\psi_1$ has an even number of $T$'s, we know $m = m_1 + m_2$ and $n = n_1 + n_2$ are both even and thus $m_1$ and $m_2$ have the same parity, and same for $n_1$ and $n_2$. Also, since by inductive hypothesis $\psi_2$ has an even number of $T$'s and $F$'s in the truth-table, we have that $m_1 + n_1$ and $m_2 + n_2$ are both even, meaning that $m_1$ and $n_1$ have the same parity, and same for $m_2$ and $n_2$. Combining this, that means that $m_1$ and $n_2$ have the same parity, and same for $m_2$ and $n_1$. Hence, $m_1 + p_2$ and $m_2 + p_1$ are both even, meaning that $\psi$ has an even number of $T$'s and $F$'s in the truth-table.

Now that we have proven the claim, we know that you cannot capture truth-functions that have an odd number of $T$'s and an odd number of $F$'s in the truth-table. Hence, $\{ \neg, \leftrightarrow \}$ is not expressively complete.