According to my studies, an argument is considered valid if and only if the conjunction of the premises implies the conclusion.
i.e $$p(p \to [q \to (r \to s)]) \to s$$ is a tautology.
One way to do this is to show that the above function can be reduced to 1 with standard simplifications
please see my working out in the attached picture. I managed to reduce this function to 1 and thus the argument should be valid, however the answers says that the argument is invalid.
If I am wrong, then why am I able to reduce this function to 1?
If I am correct, please let me know
my solution:
Consider the counter example $p=1$ and $q=r=s=0$
\begin{align} f=&p(p\to(q'+r'+s))\to s\\ =&1(1\to(0'+0'+0))\to 0\\ =&1(1\to1)\to 0\\ =&1\to0=0 \end{align}
The problem is your last step, $(p'+p)(p'+qrs')+s$ does not equivalent to $1$. Instead apply Negation law indeed $(p'+p)=1$, and by Identity law it's $(p'+qrs')+s$. \begin{align} f=&p(p\to(q'+r'+s))\to s\\ =&p'+(p'+q'+r'+s)'+s\\ =&p'+pqrs'+s\\ =&\underline{(p'+p)}(p'+qrs')+s\\ =&\underline{(1)}(p'+qrs')+s\neq1 \end{align}