there.
I was taking a look at the solution to a First Order Logic (FOL) problem I had that asked if a statement was FOL Valid/Consequent. I was not able to produce a counter-example world, and when I saw the solution, I realized that my instructor had replaced one of the predicates that took a single argument, with a predicate that took two arguments!
Is this even possible!? And if so, what governs the rules behind it?
The statement is as follows...
∃x(P(x) ^ Q(x))
∃x(P(x) ^ R(x))
Therefore... ∃x∃y(P(x) ^ P(y) ^ x≠y)
The provided counterexample is as follows.
Let... P(x) mean Small(x)
Let... Q(x) mean Cube(x)
Let... R(x) mean Left(x,a)
These being the case, and these being plugged into the original FO statements, the premises would be true and the conclusion would be false.
∃x(Small(x) ^ Cube(x))
∃x(Small(x) ^ LeftOf(x,a))
Therefore... ∃x∃y(Small(x) ^ Small(y) ^ x≠y)
I appreciate any insight that anyone may give regarding this question.
Thank you ^^~
$\begin{array}{|l}1.~\exists x ~(\mathsf{Small}(x)\wedge\mathsf{Cube}(x))\\ 2. ~\exists x~(\mathsf{Small}(x)\wedge\mathsf{Tetrahedron}(x))~\\\hline \therefore~\exists x~\exists y~(\mathsf{Small}(x)\wedge \mathsf{Small}(y)\wedge x\neq y)\end{array}$
This deduction is valid only because we implicitly accept that cubes and tetrahedrons are never the same object. A formal proof for this deduction would make use of this Analytical Consequence, $\forall x~\forall y~(\mathsf {Cube}(x)\wedge\mathsf{Tetrahedron}(y)~\to~ x\neq y)$.
However, if we replace these meaningful predicates with something where we cannot assert the same Analytical Consequence, we may no longer be able to prove the deduction.
The counter example uses predicates were that form of analytical consequence surely cannot be asserted as the only object to the left of object $a$ is in fact the only small object in that world and also a cube.
(Don't get hung up on $\mathsf{LeftOf}(x,a)$ having two arguments, as $a$ is constant, so $x$ is the only variable used in the predicate function.)