Proving that $\{p\to q, p\to \neg q\}\Rightarrow\neg p$ 
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?  
 A: 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.
A: $$\{p\to q, p\to\neg q\}\vdash \neg p$$
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.
$$\begin{align}
&(p \to q)\wedge (p\to \neg q)
\\ \Updownarrow & \qquad\text{: implication equivalence}
\\ &(\neg p\vee q)\wedge (\neg p \vee \neg q)
\\ \Updownarrow & \qquad\text{: distribution}
\\ & \neg p \vee (q\wedge \neg q) & \iff p\to (q\wedge \neg q)
\\ \Updownarrow & \qquad\text{: a contradiction}
\\ & \neg p \vee \bot
\\ \Updownarrow & \qquad\text{: disjunction's identity}
\\ & \neg p
\end{align}$$
A: 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.
