How to prove $M \models \forall x \forall y \exists z(R(x,\ y)\rightarrow R(y,\ z))$ is corret is a specific model?

Question

I have been learning Predicate Logic recently and I get the following question:

$Assumed\ F\overset{def}=\phi,\ P\overset{def}=\{R\},\ a\ model\ M\ of\ (F, P)\ is\ defined\ as\ followed:\ \\ - Universe A\overset{def}=\{a,\ b,\ c\} \\ - R^{M} \overset{def}=\{(b,\ c), (a,\ b), (c, b)\} \\ M \models \forall x \forall y \exists z(R(x,\ y)\rightarrow R(y,\ z))$

So I think it correct and reasons are like following:

$(b,\ c) \rightarrow (c,\ b), in\ which\ x\ is\ b\ and\ y\ is\ c $

$(a,\ b) \rightarrow (b,\ a), in\ which\ x\ is\ a\ and\ y\ is\ b $

$(c,\ b) \rightarrow (b,\ c), in\ which\ x\ is\ c\ and\ y\ is\ b $

I think it is correct but I don't know how to prove it in a mathematical rigor way.

if not mind, could anyone tell me the mathematical rigor way and prove it.

Thanks in advance.

Answer 1

To prove a statement is true in some model, you 'merely' have to interpret the statement relative to the model, and show that, as such, the statement is indeed true. This demonstration has to satisfy reasonable standards of a mathematical demonstration ... meaning it should be rigorous, but it does not have to be stated in formal logic, if that's what you are afraid of. Indeed, for a mathematical demonstration, it is ok to 'talk' the reader through it using English... as long as it is rigorous.

Now, it seems like you understood the basic interpretation of the sentence, which states that for every ordered pair $(x,y) \in R^M$, there should be some pair $(y,z) \in R^M$, i.e. a pair whose first element is the same as the second element of the first ordered pair.

And, since there are only three ordered pairs, you can indeed just consider each pair and see if you can find a second pair with that condition as specified, which is exactly what you did. The only small problem you have is that for pair $(a,b)$ you claimed that there is a second pair $(b,a)$, but that is not true, since $(b,a) \not \in R^M$. So, is there some other pair you can point to?

Otherwise, you proof is correct and is rigorous enough, although I would start by explicitly stating the interpretation of the logic sentence.