# Section 1.5: Problem 8 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Let $\mathbb{I}$ be the ternary connective such that $\mathbb{I}\alpha\beta\gamma$ is assigned the value $T$ iff exactly one of the formulas $\alpha$ , $\beta$ , $\gamma$ is assigned the value $T$ . Show that there are no binary connectives $\circ$ and $\triangle$ such that $\mathbb{I}\alpha\beta\gamma$ is equivalent to $(\alpha\circ\beta)\triangle\gamma$ .
Intuitively, given $\alpha$ and $\beta$ , $\mathbb{I}\alpha\beta\gamma$ is true iff $\gamma$ is true when both $\alpha$ and $\beta$ is false, or $\gamma$ is false if exactly one of $\alpha$ and $\beta$ is true. Therefore, we have three cases, when neither $\alpha$ nor $\beta$ is true, when exactly one of $\alpha$ and $\beta$ is true, and when both are true. In the first case $\mathbb{I}\alpha\beta\gamma$ results in $T$ iff $\gamma$ is assigned $T$ , in the second case iff $\gamma$ is assigned $F$ , and in the third case never. At the same time, $(\alpha\circ\beta)\triangle\gamma$ is true for some values of $\gamma$ and $(\alpha\circ\beta)$ , where we cannot distinguish the three cases for $\alpha$ and $\beta$ , i.e. $(\alpha\circ\beta)$ can be either false or true, which does not tell us in which case we are, but then the result of $\triangle$ should depend on this.
More formally, suppose that $F\circ F=\$ where $\\in\{T,F\}$ (in the usual sense in terms of evaluations). Then, $\\triangle F=(F\circ F)\triangle F=F$ and $\\triangle T=(F\circ F)\triangle T=T$ . Now, suppose that $F\circ T=\%$ where $\%\in\{F,T\}$ . Then, $\%\triangle F=(F\circ T)\triangle F=T$ and $\%\triangle T=(F\circ T)\triangle T=F$ . Finally, assume that $T\circ T=@$ where $@\in\{T,F\}$ . Then, $@\triangle F=(T\circ T)\triangle F=F$ and $@\triangle T=(T\circ T)\triangle T=F$ . Note, that $\\triangle F\neq\%\triangle F\neq@\triangle F$ and $\%\triangle T\neq\\triangle T\neq@\triangle T$ . Hence, $\\neq\%\neq@$ and $\\neq@$ , which is not possible as $\,\%,@\in\{F,T\}$ .