Predicate logic, transitivity (sort of?) I have a question. It involves 2 pictures for which I'm supposed to write a formula which is true for one, but false for the other.
The pictures can be found here on page 23 (the arrows pointing in a square)
Or you can find just the question and the squares here
My answer would be this:
$$\exists x \exists z (\forall y (Rxy \wedge Ryz)\rightarrow Rxz)$$
With $R$ being "$x$ has a relation to $y$".
Which is similar to transitivity. Would that be correct?
Thanks in advance, Rope.
P.S. Again sorry for asking so many questions, but I don't have the answers to these questions as it is a selfstudy.
 A: I think we must have :

$(∃x)(∃y)(∃z)[\quad ¬(x = y) ∧ ¬(x = z) ∧ ¬(y = z) ∧ R(x, y) ∧ R(x, z)] \land (\forall w)(¬(w = y) ∧ ¬(w = z) ∧ ¬(y = z) ∧ R(w, y) ∧ R(w, z) \rightarrow (x = w)) \quad ]$

A: I think it helps to try these things in intuitive, plain-language terms first. So for instance I would start with either of the following sentences:


*

*There is at most one $x$ that is a sender but not a receiver.

*There is an $x$ that is both a sender and a receiver.
Both of these are true for the left picture but false for the right one.
EDIT: The trouble I see with the OP proposal is that it's true whenever there exist $x$ and $z$ such that $Rxz$, because those make the consequent true and now we don't care about the antecedent. In short, the formula is true for both pictures.
EDIT2: If we want to look for something related to transitivity, we notice that the relation in the left picture is not transitive but the one in the right picture is, vacuously (these is no $x$ that plays the role of "middle element"). So why not just use the definition of transitivity? $$\forall x \forall y \forall z((Rxy \wedge Ryz) \rightarrow Rxz)$$
