$\neg P \implies \neg T$ and $P \implies \neg T$. Where do I go next?

Question

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?

Answer 1

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.

Answer 2

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.)

Answer 3

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

Answer 4

Here's another way: $$ \begin{array}{c|c|c} \# & \text{Statement} & \text{Reason}\\ \hline (1) & T & \text{Assumption} \\ (2) & \neg P \implies \neg T & \text{Given} \\ (3) & T \implies P & \text{Contrapositive of (2)} \\ (4) & P & \text{(1), (3), and Modus Ponens} \\ (5) & P \implies \neg T & \text{Given} \\ (6) & T \implies \neg P & \text{Contrapositive of (5)} \\ (7) & \neg P & \text{(1), (6), and Modus Ponens} \\ (8) & P \land \neg P & \text{(4), (7), and Conjunction} \\ (9) & \bot & \text{(8) and Inverse Law} \\ (10) & \neg T& \text{(1)-(9) and Proof By Contradiction} \\ \end{array} $$