Check if each of this sentences are true.

Question

Check whether each of this statements are true or not: $1).$ If there is a tableau $T$ for $\varphi$ where each branch of $T$ is open, then $\varphi$ is a valid formula. $2).$ If $\varphi$ is valid and $\varphi \models \psi$, then $\psi$ is also valid. $3.)$ Id $\varphi$ is valid and $\psi \models \varphi$ then $\psi$ is also valid. $4).$ If $\varphi$ is a contingency and $\psi \models \varphi$, then $\psi$ is also a contingency.

For $1).$, we can't know from the data given. Take $\varphi =p \land q$. A tableau $T$ for $\varphi$ has all of its branches open but $\varphi$ is a contingency an hence, not logically valid.

For $2.)$ This is true. From hypothesis, since $\varphi$ is logically valid, then for each interpretation $\mathcal{I}$, we have $\mathcal{I} \models \varphi$. So take any interpretation $\mathcal{I}$. We wish to prove that $\mathcal{I} \models \psi$, but this is true, since $\mathcal{I} \models \varphi$ and $\varphi \models \psi$.

For $3).$. This doesn't have to be true. Take $\varphi = \top$. Then $\psi \models \varphi$ for all $\psi$. In particular, take $\psi = p\land q$, which is a contingency. Then we have that $\psi \models \varphi$ but $\psi$ is not logically valid.

For $4).$ I believe that doesn't have to be true always. If $\varphi$ is a contingency, then we can find two interpretations $\mathcal{I}_1$ and $\mathcal{I}_2$ such that $\mathcal{I}_1 \models \varphi$ and $\mathcal{I}_2 \not \models \varphi$. Also, if $\psi \models \varphi$, then for each interpretation such that $\mathcal{I} \models \psi$, we must have that $\mathcal{I} \models \varphi$ as well. But if $\psi$ is a contradiction, then there is not interpretation such that $\mathcal{I} \models \psi$ in the first place, but $\varphi$ is still a contingency.

Are this reasonings correct? Thank you in advance!

Answer 1

Yes, you are correctly thinking about all 4 cases. A couple of notes:

• For 1), 3), and 4), the statements are meant as general statements involving any $\varphi$ and $\psi$. A such, these three statements are all false, for exactly the reasons you indicate.
• For 1), you can also just use $P$ as a counterexample
• For 2), you can put this a bit sharper by saying: Since $\varphi$ is logically valid, then for each interpretation $\mathcal{I}$, we have $\mathcal{I} \models \varphi$ (A). We also have $\varphi \models \psi$, which means that for any interpretation $\mathcal{I}$, we have that if $\mathcal{I} \models \varphi$ then $\mathcal{I} \models \psi$ (B). So, consider any interpretation $\mathcal{I}$. By (A), we get that $\mathcal{I} \models \varphi$, and so by (B) we get that $\mathcal{I} \models \psi$. So, for each interpretation $\mathcal{I}$, we have $\mathcal{I} \models \psi$, and hence $\psi$ is valid.

Otherwise, very good!