$\neg P \implies \neg T$ and $P \implies \neg T$. Where do I go next? I can't find any logic equivalence or inference rules on this. Personally, I feel that $\neg P \implies \neg T$ and $P \implies \neg T$ would mean that it follows that $\neg T$ is true regardless, and I should be able to use that fact as such in my next step. Is this proper reasoning, though?
 A: Note that $P\rightarrow T$ is equivalent to $\lnot P\lor T$. Therefore we have equivalence between $(P\rightarrow\lnot T)\land(\lnot P\rightarrow\lnot T)$ and $(\lnot P\lor\lnot T)\land(P\lor\lnot T)$.
By distributivity this can be written as $(\lnot P\land P)\lor\lnot T$, which is equivalent to $\lnot T$.
Therefore $(P\rightarrow\lnot T)\land(\lnot P\rightarrow\lnot T)$ is logically equivalent to $\lnot T$. If the former is assumed true, then so is the latter. If $T$ is not true, then so is the former.
A: I don’t know in what system you’re working, but $\neg P\to\neg T$ is equivalent to $T\to P$, and $P\to\neg T$ is equivalent to $T\to\neg P$. $(T\to P)\land(T\to\neg P)$ is equivalent to $T\to(P\land\neg P)$, which is equivalent to $\neg T\lor(P\land\neg P)$, i.e., to $\neg T\lor\bot$, where I use $\bot$ for a contradiction. (You may use F or the like.) Finally, $X\lor\bot$ is always equivalent to $X$, so the conjunction of your original implications is equivalent to $\neg T$.
(Mind you, if I wanted formal verification of what I agree is a pretty obvious conclusion, I’d probably not bother with an algebraic approach here: I’d just use a truth table to establish that
$$\Big((\neg P\to\neg T)\land(P\to\neg T)\Big)\leftrightarrow\neg T$$
is a tautology.)
A: The Wikipedia list of rules of inference lists case analysys
$$\begin{align}&\phi\lor\psi\\&\phi\to\chi\\&\underline{\psi\to\chi}\\&\chi\end{align} $$
We could make use of this with $\phi\equiv P$, $\psi\equiv \neg P$, and $\chi\equiv \neg T$, provided we knew
$$P\lor\neg P. $$
This is tertium non datur and holds for all $P$ unless you work in intuitionistic logic
A: Here's another way:
$$ \begin{array}{c|c|c}
\# & \text{Statement} & \text{Reason}\\
\hline
(1) & T & \text{Assumption} \\
(2) & \neg P \implies \neg T & \text{Given} \\
(3) & T \implies P & \text{Contrapositive of (2)} \\
(4) & P & \text{(1), (3), and Modus Ponens} \\
(5) & P \implies \neg T & \text{Given} \\
(6) & T \implies \neg P & \text{Contrapositive of (5)} \\
(7) & \neg P & \text{(1), (6), and Modus Ponens} \\
(8) & P \land \neg P & \text{(4), (7), and Conjunction} \\
(9) & \bot & \text{(8) and Inverse Law} \\
(10) & \neg T& \text{(1)-(9) and Proof By Contradiction} \\
\end{array} $$
