Logic: How to prove this argument is not valid Good day to all,
I need calcification on how to show this argument is not valid by finding a counterexample, but without using a truth table. Since there are 5 propositions I would need a 32 row truth table. It would be too time consuming to construct.
                    p
                    p v q 
                    p --> (r --> s)
                    t --> r

                   ∴ ~s --> ~t 

Keeping it short I have this in the end.

p is True
q is False
r is True
s if False
t is True

If so, how is that a counterexample? Shouldn't the argument be valid?
Thanks
This is how I worked it out:
My understanding is that for the first four premises we can`t deduce it inference rule as their value could be either true or false which the result will still give me true.
Hence we focus on the conclusion that ~t is "False" which makes ~s "False"
so that the conclusion will be true. Such that we also narrow the input of t--> r . to be True and True. Since negation of t is False.  So on and so forth until we reach the first premise.
 A: You've made a serious mistake in your produced counterexample. With $r$ being true and $s$ being false, then $r\rightarrow s$ is false, and with $p$ being true $p\rightarrow(r\rightarrow s)$ is false.
To be complete in investigating with truth table we first simplify and check which rows that are ruled out by the premisses. First of all we can eliminate $q$ as it is not used in the conclusion (in addition it doesn't rule out any possibility that the $p$ premise doesn't already rule out).
Of course every row with $P$ being false is ruled out by the first premise. This leaves us with a truth table with only $8$ rows. In addition $p\rightarrow (r\rightarrow s)$ rules out the rows where $s$ is false and $r$ is true (and $p$ is also true, but we already ruled out any row that $p$ is not) and the $t\rightarrow r$ rules out the rows where $r$ is false yet $t$ is true. This leaves us with:
$\begin{matrix}
r & s & t \\
\hline \\
0 & 0 & 0 \\
0 & 0 & 1 & \text{ruled out by } t\rightarrow r \\
0 & 1 & 0 \\
0 & 1 & 1 & \text{ruled out by } t\rightarrow r \\
1 & 0 & 0 & \text{ruled out by } p\rightarrow(r\rightarrow s) \\
1 & 0 & 1 & \text{ruled out by } p\rightarrow(r\rightarrow s) \\
1 & 1 & 0 \\
1 & 1 & 1 \\
\end{matrix}$
So we see that what is left is only rows with $s, t$ being both false, both true or $s$ being true and $t$ being false. Which all making $\neg s\rightarrow \neg t$ being true.
A: The argument is valid.
For the four premises to be all true, we must evaluate (1) $p$ as true, (2) $q$ as either true or false (we cannot infer which), (3) $r$ as false or $s$ as true (since $r\to s$ is an inference from evaluating $p$ being true), and (4) $t$ as false or $r$ as true.
In short, we must resolve to have $s$ as false or $t$ as false.   Therefore $\neg s\to \neg t$ will be evaluated as true when the premises all are so.
A: One way to check if the argument is valid is to use a tree proof generator. This tree proof generator shows the argument is valid. If it found a branch that did not close with an "x", it would output a counterexample constructed from a branch that did not close.

To try finding a counterexample manually, consider the conclusion $\neg s \to \neg t$. What truth values of $s$ and $t$ would make that conditional false? If we set $s$ false, then $\neg s$ is true. If we set $t$ true, then $\neg t$ is false. If true implies false the conditional is false.
The conclusions provides constraints for valuations we assign to the other propositions.  Those valuations have to make all of the premises true. 

Tree Proof Generator. https://www.umsu.de/trees/
