Propositional Logic Proof using I.P. or C.P or rules of inference I'm attempting to solve a proof my professor asked. We are able to use any of the rules of inference, Indirect Proof or Conditional Proof. Every time I think am making progress I run into a brick wall. Here is the question.


*

*$Q \lor (R \rightarrow S)$

*$[R \rightarrow  (R \rightarrow  S)] \rightarrow (T \lor U)$

*$(T \rightarrow Q) \land (U \rightarrow V)$

*Conclusion: $Q \lor V$


I believe the easiest solution would be to attain $(T \lor U)$ from line 2 and then use as a Constructive Dilemma with line 3 but I'm really struggling to get past the $[R \rightarrow  (R \rightarrow  S)]$ part in order to get $(T \lor U)$. If anyone can help it would be greatly appreciated.
edit* I got by the previously mentioned part, but I am now struggling to get $(R \rightarrow S)$ from line one.
Translations:


*

*"$\supset = \rightarrow$"(if...then)

*"$\bullet = \land$"(and)

*~ = $\lnot$(not)



here is my work thus far. I have been trying any and everything for the past 4 hours and I have no idea where I am going from here. 
 A: *

*[Q∨(R→S)] assumption

*{[R→(R→S)]→(T∨U)} assumption

*[(T→Q)∧(U→V)] assumption

*[~~Q∨(R→S)] 1 double negation

*[~Q→(R→S)] 4 material implication

*[~Q→((R∧R)→S)] 5 ∧ tautology

*[~Q→((R→(R→S)] 6 exportation

*[~Q→(T∨U)] 7, 2 hypothetical syllogism

*(U→V) 3 simplification (this step isn't correct... we first have to use ∧ commutativity, and then use simplification.. shall I edit this to make this explicit, or is this clear enough?).

*[~Q→(~~T∨U)] 8 double negation

*[~Q→(~T→U)] 10 material implication

*[(~Q∧~T)→U) 11 exportation

*[(~Q∧~T)→V] 12, 9 hypothetical syllogism

*[(~T∧~Q)→V] 13 ∧ commutation

*[~(T∨Q)→V] 14 De Morgan

*[~~(T∨Q)∨V] 15 material equivalence

*[(T∨Q)∨V] 16 double negation

*[T∨(Q∨V)] 17 associativity

*[(Q∨V)∨T] 18 ∨ commutativity

*(T→Q) 3 simplification

*[~~(Q∨V)∨T] 19 double negation

*[~(Q∨V)→T] 21 material implication

*[~(Q∨V)→Q] 21, 19 hypothetical syllogism

*[~~(Q∨V)∨Q] 23 material implication

*[(Q∨V)∨Q] 24 double negation

*[(V∨Q)∨Q] 25 ∨ commutativity

*[V∨(Q∨Q)] 26 associativity

*[V∨Q] 27 ∨ tautology

*[Q∨V] 28 ∨ commutativity
A: It turns out $R \rightarrow (R \rightarrow S)$ is actually equivalent to $R \rightarrow S$.  You can show this in a couple of different ways: use Material Equivalence twice, plus the associative and commutative laws for disjunction, or alternatively, use exportation.  (For the latter, though, you'd need some extra lines to justify turning $R \wedge R$ into $R$, since this doesn't appear to be one of your allowed equivalences.)
How to fit this into an overall solution, based on Indirect Proof: Suppose the desired conclusion is false, which by De Morgan and Simplification gives you both $\neg Q$ and $\neg V$.  By Disjunctive Syllogism you then obtain $R \rightarrow S$, and after applying the aforementioned equivalence followed by Modus Ponens, you arrive at $T \vee U$.  Simplifying the third premise and applying a Constructive Dilemma gets you to $Q \vee V$, a contradiction.
A: We can "formalize" Dave's answer, using Indirect Proof  : 
1) $\lnot (Q \lor V) \equiv (\lnot Q \land \lnot V)$ --- assumed [1] and using (DM)
2) $\lnot Q$ --- from 1) by (Simp)
3) $R \rightarrow S$ --- from premise 1. and 2) by (DS)
4) $(R \land R) \rightarrow S$ --- from 3) by (Taut) 
5) $R→(R→S)$ --- from 4) by (Exp)
6) $T \lor U$ --- from premise 2. and 5) by (MP)
7) $Q \lor V$ --- from 6) and premise 3. by (CD)
but 7) contradicts 1); thus we have :

9) $Q \lor V$ --- by Indirect Proof, discharging [1].

Conclusion :


$Q∨(R→S), [R→(R→S)]→(T∨U), (T→Q)∧(U→V) \vdash Q \lor V$.


