TITLE: Is equality of formulas with floor rounding or integer division decidable?
QUESTION [8 upvotes]: As far as I know, formulae involving rationals and basic arithmetic ($+$, $-$, $\cdot$ and $/$) have decidable equality. Is this still the case if we add floor rounding (or integer division)?
Define a "basic formula" by the following grammar (in Backus-Naur form):
$$
\begin{matrix}
e &= &q \\
  &| &v \\
  &| &e + e \\
  &| &e - e \\
  &| &e \cdot e \\
  &| &e / e \\
  \end{matrix}
$$
Here, $q$ stands for rational literals and $v$ for a set of variables.
A valuation is the assignment of a rational number to every variable.
The semantics of such a formula for a given valuation is ordinary arithmetics. Two formulae are semantically equal if they are equal for every allowed valuation. (Division by 0 is not allowed, of course.)
As far as I understand, this notion of equality is decidable. Given two formulas, we essentially multiply by all denominators, apply the distributive law everywhere and then compare summands. We've basically reduced each expression to a normal form where we can compare.
Now, let us add the floor function, which rounds down to the nearest integer. We extend the above definition by:
$$
\begin{matrix}
e &= &\dots \\
  &| &\lfloor e \rfloor \\
  \end{matrix}
$$
Is equality still decidable here? It's not so obvious to me that a well-behaved normal form even exists, since the floor function satisfies some equations such as $\lfloor \lfloor x \rfloor + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$, Hermite's identities (e.g. $\lfloor x\rfloor + \lfloor x+1/3\rfloor + \lfloor x+2/3\rfloor = \lfloor 3x\rfloor$), and so on.
Is there still some simple enough algorithm that decides whether two such formulae are equal?
Note: I could have added integer division instead of the floor function as well, but I believe that this wouldn't make a difference.

Feel free to add or suggest tags, I'm not so sure whether the question is categorized well.

REPLY [11 votes]: Equality of formulas is undecidable. To prove this it is enough to show that an algorithm for equality of formulas would enable us to determine whether or not a multivariate polynomial with integer coefficients vanishes at at least one integer point, because the latter condition is undecidable according to the negative solution to Hilbert's Tenth Problem.
But the polynomial $f(x)$, where $x$ is a tuple of variables and $f$ has integer coefficients, does not vanish at any integer point if and only if the following equation holds at all integer points $x$: 
$$\Bigl\lfloor\dfrac{1}{1+f(x)^2}\Bigr\rfloor=0.$$
I believe this settles it. However it would be nice to know a simple elegant decidability result for formulas excluding division.