Proof by induction of inadequacy of a propositional connective I have the following truth table of a newly defined logical operator and have to prove its functional incompleteness via structural induction.
My idea is that that you cannot express the always true formula $\top$ in terms of this operator.
I just don't know how to go about the proof.
Here is the truth table of the new 3-input Operator:
$$\begin{array}{|c|c|c|c|} 
 X & Y & Z & <X,Y,Z> \\ \hline
0 & 0 & 0 & 0\\ \hline
0 & 0 & 1 & 0\\ \hline
0 & 1 & 0 & 0\\ \hline
0 & 1 & 1 & 1\\ \hline
1 & 0 & 0 & 1\\ \hline
1 & 0 & 1& 0 \\ \hline
1 & 1 & 0 & 0\\ \hline
1 & 1 & 1 & 0\\ \hline
\end{array}$$
 A: We can use structural induction to give a careful proof that any term we can build using only the function symbol $<-,-,->$ and the variable symbol $x$ represents either the identity function or the constant $0$ function.
Base case $x$: Upon substituting $1$ for $x$, we get $1$. Upon substituting $0$, we get $0$. Therefore $x$ represents the identity function.
Inductive case: Consider a term $f(x)$ form $<A(x),B(x),C(x)>$, where we know by inductive assumption that each of $A(x)$, $B(x)$ and $C(x)$ represents either the identity function or the constant 0 function.
We perform a case analysis, tabulating each of these eight possibilities:
A(x) | B(x) | C(x) | x | f(x)
-----|------|------|---|-----
 x   |  x   |  x   | 1 |  0
     |      |      | 0 |  0   (const. 0)
-----|------|------|---|-----
 x   |  x   |  0   | 1 |  0
     |      |      | 0 |  0   (const. 0)
-----|------|------|---|-----
 x   |  0   |  x   | 1 |  0
     |      |      | 0 |  0   (const. 0)
-----|------|------|---|-----
 x   |  0   |  0   | 1 |  1
     |      |      | 0 |  0   (identity)
-----|------|------|---|-----
 0   |  x   |  x   | 1 |  1
     |      |      | 0 |  0   (identity)
-----|------|------|---|-----
 0   |  x   |  0   | 1 |  0
     |      |      | 0 |  0   (const. 0)
-----|------|------|---|-----
 0   |  0   |  x   | 1 |  0
     |      |      | 0 |  0   (const. 0)
-----|------|------|---|-----
 0   |  0   |  0   | 1 |  0
     |      |      | 0 |  0   (const. 0)

We see that in each of these cases, $<A(x),B(x),C(x)>$ represents either the identity function or the constant $0$ function. By the principle of structural induction, any term built using only the function symbol $<-,-,->$ and the variable symbol $x$ represents either the identity function or the constant $0$ function.
Consequently, we cannot express negation or the constant $1$ function using only $<-,-,->$, and so the set $\{ <-,-,-> \}$ is not functionally complete.
A: $<X,Y,Z>$ or not-$<X,Y,Z>$
has a truth table of all 1's.  
