Why is $\forall xO(x,c), \forall xO(c,x) \therefore \forall xO(x,x)$ an invalid reasoning?
This exercise appears in P.D. Magnus. "forallX: an Introduction to Formal Logic" (p. 268, exercise D. 7).
I think to satisfy both universal quantifiers, predicate $O$ must be symmetric, so I cannot find an interpretation that makes $\forall xO(x,x)$ false. On the surface, it seems a valid reasoning (considering a non-empty universe).
What's happening here?
On
On the surface, it seems a valid reasoning (considering a non-empty universe). What's happening, here ?
Since there are already some answers that give neat concrete counterexamples to the claim, I will try to give you an intuitive picture of why it is not a valid inference.
The statement $\forall x O(x,c)$ says that $O$ is true for all $x$ when $x$ is on the left, and the statement $\forall x O(c,x)$ says that $O$ is true for all $x$ when $x$ is on the right. Although it appears to be all about $x$, this is really more of a statement about $c$. It simply says that $c$ makes $O$ true whenever it appears on the left or the right. $x$ is just a dummy variable we use to write this statement formally—it is standing in for anything in the domain.
Once you realise this, it should be fairly obvious that $O(x,x)$ is not necessarily true for all $x$. If $x \neq c$ then we have no guarantee that $O$ is true in this case, as our original statement only refers to situations involving $c$.
On
Remember that validity means the argument is valid under all interpretations, so to refute it, it suffices to come up with one counter example.
To go at your intuition, it often helps to think of a more informal interpretation. Think of a classroom setting, where $O$ stands for "sees" and $c$ refers to the teacher. Everyone can see the teacher, and the teacher sees everyone, including the himself in a reflection in the window, so $\forall x O(c,x)$ and $\forall x O(x,c)$ are true. But the students sitting in the front row can't see themselves or the students behind them, so $\forall x O(x,x)$ is false.
If you want to write it down in formal notation, to simplify let's say there is just one student in the classroom, then we have
$D = \{Alice, Mrs. Smith\},\\
F(c) = Mrs. Smith, F(O) = \{\langle Mrs. Smith, Alice \rangle, \langle Mrs. Smith, Mrs. Smith \rangle\}$
as a counter model.
Since there is at least one interpretation of the non-logical symbols where the premises are true but the conclusion is false, the argument can't be valid.
As computed here, let the domain be $\{0,1\}$, let $c=0$, and let $O$ be given by $\{(0,0), (0,1), (1,0)\}$, so that $\exists z\lnot O(z,z)$, namely, $z=1$.