How to prove that the following statement doesn't hold? We have a statement:
$$p \implies r \wedge t, t \vee s \implies \neg q, \neg s  \vee \neg r, \neg t \implies s \models \neg p \wedge \neg q$$
The statement should not hold. My first question is, is there a fast way to see if a statement does not hold or if it holds, to know what strategy to use to prove it does/doesn't.
My try of proof after I got the info that it doesn't hold:
So if the upper does not hold, then the following should hold:
$$p \implies r \wedge t, t \vee s \implies \neg q, \neg s  \vee \neg r, \neg t \implies s \models p \vee q$$
Now we can prove this by contradiction:
$$p \implies r \wedge t, t \vee s \implies \neg q, \neg s  \vee \neg r, \neg t \implies s , \neg(p \vee q) \models 0$$
So we get the following propositions:
$C_1 :p \implies r \wedge t \equiv (\neg p \vee r) \wedge (\neg p \vee t)\\$
$C_2: t \vee s \implies \neg q \\$
$C_3: \neg s  \vee \neg r \\$
$C_4: \neg t \implies s \equiv t \vee s$
$C_5: \neg(p \vee q) \equiv \neg p \wedge \neg q$
$C_6: \neg p \quad$ (from $C_5$)
$C_7: \neg q \quad$ (from $C_5$)
$\vdots$
$C_n: 0$
However I do not know how to continue to get to the result which is $0$ and to finish the proof.
 A: Regarding your first question: there is of course always a full truth-table that you can work out, but in this case there are $5$ variables, and so that would work out to $32$ rows. Hence, I suppose, your question as to whether there is a faster method. And yes, there is a faster method for this: The short truth-table method.
Here is how it works.
To test for consequence, you set the implying statements to True, and the implied statement to False. Then see if this assignment truth-values forces the truth-value of any other component sentence, and keep doing this until either all sentences, component sentences, and variables have been assigned a truth-value without contradiction, or until you reach some contradiction. If the former, you have apparently found a counterexample, and hence the implication does not hold. If the latter, then apparently no counterexample is possible, and hence the logical implication does hold.
OK, so here goes. First step:
\begin{array}{ccccc}
p \to r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p \wedge \neg q\\
T&T&T&T&F
\end{array}
Now, does this force any of the other statements? For example, does the fact that $p \to r \lor t$ is True force any of its component parts to be True or False?  No. And, as it so happens, the same goes for all the other statements. OK, so we immediately run into a spot of trouble here, since we have neither assigned a truth-value to all component sentences, nor have we reached a contradiction. But, in that case, we can make a choice as to how to proceed. For example, let's focus on the last statement, and note that there are two ways to make that statement False:
We either set $\neg p$ to False:
\begin{array}{ccccccc}
p \to r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&T&T&F&F
\end{array}
or we set $\neg q$ to False:
\begin{array}{ccccccc}
p \to r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&T&T&&F&F
\end{array}
Of course, we could set both $\neg p$ and $\neg q$ to False, but notice that that option is covered by both of these two possibilities, and since we want our work to be as minimal as possible, let's leave it at what we have.
Now, let's work with the second option, and see what happens. Since $\neg q$ is False, $q$ needs to be True:
\begin{array}{ccccccc}
p \to r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&T&T&&F&FT
\end{array}
And since we're effectively trying to create a row in a truth-table that would represent a counter-example, this means that $q$ has to be True everywhere else:
\begin{array}{cccrccccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&&T&\color{red}{T}&T&T&&F&FT
\end{array}
and that forces $\neg q$ to be False:
\begin{array}{ccccccccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&&T&\color{red}{F}T&T&T&&F&FT
\end{array}
OK, now note that if $\neg q$ is False, then in order for $t \vee s \to \neg q$ to be True, it has to be the case that $t \lor s$ is False:
\begin{array}{ccccccccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&\color{red}{F}&T&FT&T&T&&F&FT
\end{array}
And the only way for that to happen, is if both $s$ and $t$ are False:
\begin{array}{ccccccccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&\color{red}{F}F\color{red}{F}&T&FT&T&T&&F&FT
\end{array}
Again, copying those values elsewhere:
\begin{array}{cccccccrccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&FFF&T&FT&T&\color{red}{F}T\color{red}{F}&&F&FT
\end{array}
and that means $\neg t$ is True:
\begin{array}{ccccccccccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&FFF&T&FT&T&\color{red}{T}FTF&&F&FT
\end{array}
OK, but now we have reached a contradiction: we have a true conditional, with a True antecedent, but a False consequent:
\begin{array}{ccccccccccc}
p \to r \wedge t & t \vee s & \to & \neg q & \neg s  \vee \neg r & \neg t \color{red}{\to} s & \neg p & \wedge & \neg q\\
T&FFF&T&FT&T&\color{red}{T}F\color{red}{TF}&&F&FT
\end{array}
OK, so going back to our earlier choice point: trying to make $\neg p \land \neg q$ to False by setting $\neg q$ to False apparently doesn;t work: no such counterexample is possible. OK, but as we pointed out, we could also try and set $\neg p$ to False:
\begin{array}{cccclcc}
p \to r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&T&T&F&F
\end{array}
Let's see what happens now. First, with $\neg p$ being False, $p$ itself should be True:
\begin{array}{ccccccc}
p \to r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&T&T&F\color{red}{T}&F
\end{array}
Copy the value of $p$:
\begin{array}{ccccccccc}
p & \to & r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
\color{red}{T}&T&&T&T&T&FT&F
\end{array}
Now, with $p$ being True, the only way for for $p \to  r \land t$ to be True is for $r \land t$ to be True:
\begin{array}{ccccccccc}
p & \to & r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&\color{red}{T}&T&T&T&FT&F
\end{array}
which means that both $r$ and $t$ have to be True:
\begin{array}{ccccccccc}
p & \to & r \wedge t & t \vee s \to \neg q & \neg s  \vee \neg r & \neg t \to s & \neg p & \wedge & \neg q\\
T&T&\color{red}{T}T\color{red}{T}&T&T&T&FT&F
\end{array}
and let's copy their values elsewhere:
\begin{array}{ccclccccrrcccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&\color{red}{T}&T&&&T&\color{red}{T}&\color{red}{T}&T&&FT&F
\end{array}
This forces both $\neg r$ and $\neg t$ to be False:
\begin{array}{ccclcccccccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&T&T&&&T&\color{red}{F}T&\color{red}{F}T&T&&FT&F
\end{array}
while with $t$ being True, $t \lor s$ will have to be True as well:
\begin{array}{ccclcccccccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&T\color{red}{T}&T&&&T&FT&FT&T&&FT&F
\end{array}
OK, but then for $\neg s \lor \neg r$ to be true, it has to be the case that $\neg s$ is True:
\begin{array}{ccclcclcccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&TT&T&&\color{red}{T}&T&FT&FT&T&&FT&F
\end{array}
meaning that $s$ itself has to be False:
\begin{array}{ccclccccccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&TT&T&&T\color{red}{F}&T&FT&FT&T&&FT&F
\end{array}
And copy:
\begin{array}{cccccccccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&TT\color{red}{F}&T&&TF&T&FT&FT&T&\color{red}{F}&FT&F
\end{array}
OK, almost there!  Now, with $t \lor s$ being True, the only way for $t \vee s  \to  \neg q$ to be True is if $\neg q$ is True:
\begin{array}{ccccclccccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&TTF&T&\color{red}{T}&TF&T&FT&FT&T&F&FT&F
\end{array}
meaning that $q$ itself has to be False:
\begin{array}{cccccccccccc}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&TTF&T&T\color{red}{F}&TF&T&FT&FT&T&F&FT&F
\end{array}
So then copy $q$:
\begin{array}{ccccccccccccr}
p & \to & r \wedge t & t \vee s & \to & \neg q & \neg s  & \vee & \neg r & \neg t & \to & s & \neg p & \wedge & \neg q\\
T&T&TTT&TTF&T&TF&TF&T&FT&FT&T&F&FT&F&\color{red}{F}
\end{array}
making $\neg q$ True.
OK! We've done it! This time, we assigned truth-values to all component statements, but didn't encounter any contradiction. So, we've succesfully managed to create a counterexample, thus demonstrating that the logical implication does not hold.
OK, well, that was a long explanation, but note that in practice, you would of course assign all these truth-values in succession, and you would work them out in a single line (or maybe two, if you have a choice point). And, as long as the truth-values are forced, this goes very quickly. Indeed, when doing this on paper before creating this post, it took me about 30 seconds to find the counterexample: 20 seconds to write down the statements, and 10 seconds to work out the truth-values and find the actual counterexample.
A: Let $p, r, t$ be true while $s,q$ are false.
