 # I am trying to figure out how to use symbolic logic to represent the following theorem from linear a Simone Werner 2022-05-21 Answered
I am trying to figure out how to use symbolic logic to represent the following theorem from linear algebra.
The theorem states that, "A system of linear equations has zero, one or infinitely many solutions. There are no other possibilities."
Initially, I thought it would have the form
$p\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(q\vee r\vee s\right)$
but this clearly does not include the "There are no other possibilities" part.
How do I represent the theorem using symbolic logic?
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The expression with $q$, $r$, and $s$ means we have to define each of these expressions separately. It also obscures important facts that we already knew before we began to consider this particular theorem. For example, if a system has exactly one solution, it does not have exactly two solutions.
Instead, let $S\left(\varphi \right)$ denote the solution set of an arbitrary system $\varphi$. Let $p\left(\varphi \right)$ denote the statement that $\varphi$ is a system of linear equations. The theorem states that
$p\left(\varphi \right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|S\left(\varphi \right)|\in \left\{0,1,|\mathbb{R}|\right\}.$
This tells us there is no other possibility; for example, $|S\left(\varphi \right)|=2$ is not possible, because $2\notin \left\{0,1,|\mathbb{R}|\right\}.$
I used $|\mathbb{R}|$ to represent the cardinality of the number of solutions, since (I think) the point of the theorem was that if there is more than one solution, the solutions can be parameterized by real numbers (or complex numbers if that's the field you're using). You can simplify this a little further if you have a nice symbol to represent the cardinality $|\mathbb{R}|.$ In any case, the third element of the set (in addition to 0 and 1) should be whatever you consider to be "infinitely many."