Natural deduction - Logic and Proof Given the premise P ∨ ¬ Q  by natural deduction prove (P → Q) → ((¬ P → Q) → Q)
I am trying to prove this using a Case Proof in Fitch
For my initial subproof, would I be correct in assuming P ∨  Q. I want to deduce implication of P → Q but I am not sure which rule to site. Do I have to break P ∨  Q into Cases of P ∨ Q and cite by elim before I can conclude P → Q. 
I think the gap here is how to get from ∨ to → .
Thanks
 A: 
"For my initial subproof, would I be correct in assuming $P \lor Q$?"

This is legal, but I don't see how it could help.

"I want to deduce implication of $P \to Q$".

You can't infer this from $P\lor \neg Q$, nor from $P\lor Q$.

The formula $(P \to Q) \to ((\neg P \to Q) \to Q)$ is a known tautology, therefore you don't need any premises to prove it, (though premises might help shorten the proof). Below I give a proof of this without premises and you can work your way towards shortening it using the given premise.

A: I assume the logic under consideration is classical propositional logic. Since the classical derivability relation is monotone ($S \vdash \varphi, S \subseteq S' \Rightarrow S' \vdash \varphi$) and the empty set is included in every set it, suffices to show that the conclusion is derivable from the empty set   
This is easy: 
1(1) $P\rightarrow Q$,       Assumption
2(2) $\neg P \rightarrow  Q$, Assumption
3(3) $\neg Q$,                Assumption
4(1,3) $\neg P$, MT 1,3
5(1,2,3) Q, $\rightarrow$-elim 2, 4
6(1,2,3) $Q \wedge \neg Q$, $\wedge$-intro 3, 5
7(1,2) $Q$, RAA, 3, 6
8(1)  $(\neg P \rightarrow  Q) \rightarrow Q$, $\rightarrow$-intro 2, 7
9( ) $(P\rightarrow Q) \rightarrow ((\neg P \rightarrow  Q) \rightarrow Q)$, $\rightarrow$-intro 1, 8
A: Another approach:


*

*$P\lor \neg Q $  (Premise, 2 cases)

*$P\implies Q$  (Premise)

*$\neg P \implies Q$  (Premise)

*$P$  (Premise, Case 1)

*$Q$  (Detach, 2, 4)

*$P\implies Q$  (Conclusion, Case 1)

*$\neg Q$ (Premise, Case 2)

*$\neg Q \implies \neg P$ (Contrapositive, 2)

*$\neg P$  (Detach, 8, 7)

*$Q$ (Detach, 3, 9)

*$\neg Q \implies Q$  (Conclusion, Case 2)

*$P \implies Q \implies Q$  (Cases, 1, 6, 11)

*$P\implies Q \implies [\neg P \implies Q \implies Q]$ (Conclusion, 2)

*$P\lor \neg Q \implies [P\implies Q \implies [\neg P \implies Q\implies Q]$  (Conclusion, 1)
A: $\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}$Yet another approach.  An intuionistic proof using disjunction elimination on the premise
$$\fitch{~~1.~P\vee\neg Q\hspace{25.5ex}\textsf{Premise}}{\fitch{~~2.~P\to Q\hspace{22.5ex}\textsf{Assumption}}{\fitch{~~3.~\neg P\to Q\hspace{18ex}\textsf{Assumption}}{\fitch{~~4.~\neg Q\hspace{20ex}\textsf{Assumption}}{\fitch{~~5.~ P\hspace{18ex}\textsf{Assumption}}{~~6.~Q\hspace{17.5ex}{\to}\textsf{Elimination }5,2\\~~7.~\bot\hspace{18ex}{\neg}\textsf{Elimination }6,4}\\~~8.~\neg P\hspace{20ex}{\neg}\textsf{Introduction }5{-}6\\~~9.~Q\hspace{21ex}{\to}\textsf{Elimination }8,3}\\10.~\neg Q\to Q\hspace{18ex}{\to}\textsf{Introduction }4{-}9\\11.~Q\hspace{25ex}{\vee}\textsf{Elimination }1,3,10}\\12.~(\neg P\to Q)\to Q\hspace{15ex}{\to}\textsf{Introduction }3{-}11}\\13.~(P\to Q)\to((\neg P\to Q)\to Q)\hspace{4ex}{\to}\textsf{Introduction }2{-}12}$$
