if s ⊨ t and s∧¬t ⊨ ¬u then s ⊨ u. T or F? Logical consequences. I need to determine if the following is true (if false, I need to give a counterexample):
If $s \vDash t$ and $s \wedge \neg t \vDash \neg u$ then $s \vDash u$.
As far as I can tell, if $s \vDash t$, then $s \wedge \neg t$ is always false. But then what does that say about $u$ (or $\neg u$ for that matter)?
 A: Right. If $s \vDash t$ then you can never have $s \land \neg t$, and so $s \land \neg t \vDash \neg u$ will always be true, no matter what $u$ is. So, it should be fairly easy to pick for $u$ something such that $s \not \vDash u$, and hence the claim is False. 
Now, for a counterexample, make sure you pick statements for $s$, $t$, and $u$ (rather than truth-values), and also make sure that $s$ logically implies $t$ (since that's what $\vDash$ means ... So don't just pick any two true statements for $s$ and $t$ ... They may not imply each other).
Here, for example, is an explicit counterexample:
Pick $s = t = P$ for some atomic statement $P$, and pick $u = \neg P$. Clearly $P \vDash P$ and $P \land \neg P \vDash \neg \neg P$, but $P \not \vDash \neg P$.
A: Let $s$ be the statement $1 < 2$.  Let $t$ be the statement $2 < 3$. And let $u$ be $1 > 2$.  
A: If you wanted to find a counterexample, then the if part needs to be true and the then part needs to be false. The then part itself is an implication which can only be false if $s=T$ and $u=F$. Since the if part needs to be true and since the if part is formed by a conjunction, then the first implication must also be true. Since $s=T$, the only way the implication can be true is for $t=T$ as well. Thus, if there is a counterexample, then the assignment $s,t =T$ and $u=F$ is the only assignment that produces one. You can check that this indeed a counterexample.
A: Let $s$ and $t$ be true while $u$ be false.   That is all you need.
Then $\top\models \top$ and $\top\wedge\neg\top\models \neg\bot$ (vacuously), but $\top\not\models \bot$. 
There is an interpretation where antecedents are true while the conclusion is false, so the argument is invalid.
A: Try having $s$ and $t$ as true while $u$ is false. You can find this counterexample using the method of analytic tableaux.
What you need to do is check whether $$((s\to t)\land ((s\land\lnot t)\to \lnot u))\to (s\to u)$$ is a tautology.
