Proving that $\{p\to q, p\to \neg q\}\Rightarrow\neg p$

Question

Prove the following: $\{p\to q, p\to \neg q\}\Rightarrow\neg p$, that is, prove that $\neg p$ is a tautological consequence of $\{p\to q, p\to \neg q\}$

(Note: I write $0,1$ instead of $F,T$.)

Suppose the premises are true and the conclusion is false, so:

$p=1,p\to q=1,p\to \neg q=1$

Place $p=1$ in premises: $1\to q=1$ so $q=1$.

Place $p=q=1$ in $p\to \neg q=1$ and we get: $1\to0=1$ contradiction.

Thus the conclusion must be true.

Is my approach correct?

Answer 1

$$\{p\to q, p\to\neg q\}\vdash \neg p$$

We know $q$ is true whenever $p$ is true and $q$ is false whenever $p$ is true, so that means that $q$ is simultaneously true and false whenever $p$ is true.   However, that is contradictory, so thus we have proof that $p$ can never be true, and therefore must be false.

$$\begin{align} &(p \to q)\wedge (p\to \neg q) \\ \Updownarrow & \qquad\text{: implication equivalence} \\ &(\neg p\vee q)\wedge (\neg p \vee \neg q) \\ \Updownarrow & \qquad\text{: distribution} \\ & \neg p \vee (q\wedge \neg q) & \iff p\to (q\wedge \neg q) \\ \Updownarrow & \qquad\text{: a contradiction} \\ & \neg p \vee \bot \\ \Updownarrow & \qquad\text{: disjunction's identity} \\ & \neg p \end{align}$$

Answer 2

We know that either q or ~q.

If p, then (p and ~q) or (p and q).

In other words,

If ~[(p and ~q) or (p and q)], then ~p.

Thus:

If ~(p and ~q) and ~(p and q), then ~p.

That's:

If p => q and p => ~q, then ~p.

Answer 3

Recall $\lnot p$ means $p \to \bot$ and $p \to \lnot q$ means $p \to (q \to \bot)$.

So: Assume $p$. Then, from $p \to q$, we have $q$. Also, from $p \to (q \to \bot)$, we have $q \to \bot$. Finally, from $q$ and $q \to \bot$, we have $\bot$. So we have $p \to \bot$, as desired. The only inference rule used is modus ponens.

This also shows that the implication in question is constructively valid. In particular does not rely on assuming $p \lor \lnot p$, $q \lor \lnot q$, or any similar version of the law of the excluded middle.