Proof P by premises $(\neg P\vee Q)\to P$ I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:
$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.
The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?
 A: First of all ,you should know that implication connector does not have disribution property on right hand side.
$(\neg P \lor Q)\rightarrow P$ is equal to $(\neg P \rightarrow P)\land (Q\rightarrow P)$
$(\neg P \rightarrow P)\land (Q\rightarrow P)$ $\equiv$ $( P \lor P)\land (\neg Q\lor P)$
So, $P \land (\neg Q\lor P)$ , by elimination
$\therefore P$
A: Hint: using Boolean algebra,
$$\overline{\left( \overline{p} + q \right)} + p = p \overline{q} + p = p \,(\overline{q} + 1) = p$$
A: 
I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:
$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.
The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?

Indeed, that is not a fundamental inference in Natural Deduction.  Well, sure, you may derive it, but it is not really that efficient a strategy in the system.
To derive a consequent from its conditional, you would either (1) derive the antecedent so to use conditional elimination, or (2) show that a contradition would be derived under an assumption of the negation of that consequent, so to use reduction to absurdity.  Strategy (1) cannot be done here, thus you should attempt strategy (2).
$$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{~~1.~(\lnot p\lor q)\to p\hspace{2ex}\textsf{Premise}}{\fitch{~~2.~\lnot p\hspace{9ex}\textsf{Assumption}}{~~3.~\\~~4.~\\~~5.~\bot\hspace{10ex}\textsf{Negation Elimination (?,?)}}\\~~6.~\lnot\lnot p\hspace{11ex}\textsf{Negation Introduction (2-5)}\\~~7.~p\hspace{14ex}\textsf{Double Negation Elimination (6)}}$$
So, determine what you will be contradicting, and a way you would derive that.
A: *

*Suuppose ( for refutation) that  P is false.


*It means that the true premise  $(\neg P \lor Q) \rightarrow P$ has a false consequent.


*The only way for a true conditional with a false consequent to be true is to have a false antecedent ( by the truth table of the " if ... then " operator). So $(\neg P \lor Q)$ has to be false.


*By DeMorgan's , the negation of a disjunction amounts to a conjoint negation of its disjuncts. So $\neg ( \neg P \lor Q) \equiv ( \neg \neg   P \& ~Q )$


*Appliy $\&$ Elim to get $\neg \neg   P$


*The hypothesis $\neg P$ led you to the negation of this hypothesis, namely $\neg \neg   P$.


*By Neg Intro Rule, you're entitled to conclude that the hypothesis was false, wich means that $\neg \neg   P$ is true, and ( by Double Negation ) that P is true.
