# How to prove if these statements involving logical consequences and tautologies are true?

Am struggling a lot with this particular type of question:

Are the following statements true for all propositional terms $s, t, u$? In each case briefly justify your answer, either by a proof or by giving a counterexample.

(i) If $s \vDash t → s$ and $¬s \vDash s$ then $t → s$ is a tautology.

(ii) If $s \vDash t → ¬s$ then $t \vDash ¬s$.

(iii) If $t ∨ u \vDash s$ and $t ∨ ¬u \vDash s$ then $s$ is a tautology.

Here is how I've been going about it so far:

(i) Suppose $s \vDash t \to s$ and $\neg s \vDash s$.

First, let $v$ be a valuation s.t. $v(s) = T$. Then $v(t\to s)=T$, since $s \vDash t \to s$.

Now, consider a valuation $w$ s.t. $w(s)=F$. Then $s \vDash t \to s$ tells us nothing, but we have $w(\neg s)=T$ and so $w(s)=T$ since $\neg s \vDash s$. This is clearly a contradiction, but then was does that say about $t \to s$? Does this mean that "$s \vDash t → s$ and $¬s \vDash s$" will always be false, in which case the entire statement (i) is true?

(ii) If $s \vDash t \to \neg s$, then for every valuation $v$ s.t. $v(s)=T, v(t \to \neg s)=T$, and so we must have $v(t)=F$, in which case $t \vDash \neg s$. If on the other hand, we have a valuation $w$ s.t. $w(s)=F$, then $w(\neg s)=T$. It must then be the case that $t \vDash \neg s$ is true regardless of whether $w(t)$ is $T$ or $F$. So the statement is true?

(iii) Suppose $t ∨ u \vDash s$ and $t ∨ ¬u \vDash s$. No matter whether $v(u)$ is true or false, we will have $v(s) = T$, because either $v(t \vee u)=T$ or $v(t \vee \neg u)=T$. So $s$ is a tautology. But I'm totally ignoring $t$ is this case. Is that wrong?

Any help regarding any of the 3 statements would be much appreciated.

• Sorry, that's a typo. t and s should be swapped. I'll change it right away Commented May 23, 2017 at 19:03
• For (i), it seems to me that $\lnot s \vDash s$ simply means that $s$ is always T. Thus, $t \to s$ is always T. Commented May 23, 2017 at 20:19
• Hmm, yeah, that makes sense to me. But I do wonder if ¬s ⊨ s can actually ever be true? Is there an explicit s for which it is true? Or is that completely irrelevant? Commented May 23, 2017 at 21:30
• But I have doubt about your expression "propositional term"; if $s$ is a formula, then it can be a tautology. If instead it is a propositional letter, this is impossible: to every prop letter we can always choose to assign the value T or F. Thus, for a prop letter $p$, we cannot have $\lnot p \vDash p$. Commented May 24, 2017 at 6:04

i). $\neg s \vDash s$ means that $s$ is a tautology, for if it were possible for $s$ to be false, then in that case $\neg s$ is true, and $s$ is false, and so $\neg s \vDash s$ would not hold. Also, if $s$ is a tautology, then $t \rightarrow s$ is a tautology as well. So, i) is True. The $s \vDash t \rightarrow s$ is not even needed.
ii) You got the right idea, but the execution is a little off: You can't conclude that $t \vDash \neg s$ just on the basis of a particular valuation. So, instead, try to prove that for any valuation that sets $t$ to true, $\neg s$ will have to be true as well, given $s \vDash t \rightarrow \neg s$. Put differently: show that any valuation that sets $t$ to true and $\neg s$ to galse will contradict $s \vDash t \rightarrow \neg s$
iii) Correct! Regardless of what any valuation does to $t$, it must set either $u$ or $\neg u$ to true, and therefore it must set either $t \lor u$ to true or $t \lor \neg u$ to true (or both). Hence, any valuation must set $s$ to true, and thus $s$ is a tautology.