Richardson's Theorem states that (quoting from Wolfram MathWorld):
Let $\mathcal{R}$ be the class of expressions generated by
- The rational numbers and the two real numbers $\pi$ and $\ln 2$,
- The variable $x$,
- The operations of addition, multiplication, and composition, and
- The sine, exponential, and absolute value functions.
Then if $E \in \mathcal{R}$, the predicate "$E=0$" is recursively undecidable.
I'm not sure if I understand this correctly. It seems that I can select for example:
- $\sin(x) \in \mathcal{R}$
- $1 - \sin(x)^2 - \cos(x)^2 \in \mathcal{R}$
For the first choice, it is clear that $E\equiv 0$ is false while for the second choice, it is clear that $E\equiv 0$ is true. Both expressions are contained in the set $\mathcal{R}$, so I don't understand how the theorem can state that these are undecidable.