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 $\$,\%,@\in\{F,T\}$ .