Using a calculation to show this formula is a tautology In a chapter regarding strenghtening and weakening of propositions, the following is asked:
Show with a calculation that the following is a tautology:
$$( R \land (P \Rightarrow Q)) \Rightarrow ((R \Rightarrow P) \Rightarrow Q) $$
I'm not sure I'm on the right track with my solution or if I'm missing steps. Here goes:
To show this, it is enough to show that 
$$( R \land (P \Rightarrow Q)) \vDash ((R \Rightarrow P) \Rightarrow Q) $$
I begin by rewrting the right hand side.
1) Implication, twice:
$$ \neg (\neg R \lor P) \lor Q $$
2) De morgan
$$ (\neg \neg R \land \neg P) \lor Q$$
3) Double negation
$$ (R \land \neg P) \lor Q$$
4) Distribution and implication
$$ (Q \lor R) \land (P \Rightarrow Q) $$
Now, by the standard weakening rules I know that $ R \vDash Q \lor R$
So this concludes the proof that
$$( R \land (P \Rightarrow Q)) \vDash ((R \Rightarrow P) \Rightarrow Q) $$
and thus ive shown $$( R \land (P \Rightarrow Q)) \Rightarrow ((R \Rightarrow P) \Rightarrow Q) $$ is a tautology.
Now, does this make any sense :) Especially the weakening part. Can I use that rule like this?
Text used: http://www.amazon.com/Logical-Reasoning-A-First-Course/dp/095430067X in which $P \vDash Q$ is specified as meaning 'P is a stronger proposition than Q'. 
False being the strongest, true being the weakest: $$ False \vDash P \land Q \vDash P \vDash P \lor Q \vDash True$$
 A: $R \Rightarrow (Q \lor R)$ is certainly a tautology, so you can use it. 
Put another way, you can assume $R \land (P \Rightarrow Q)$. So now you have $R$ as well as $P \Rightarrow Q$. From the above tautology and $R$, you have $Q \lor R$. Hence, you have $(Q \lor R) \land (P \Rightarrow Q)$, which you've already established as being equivalent to $((R \Rightarrow P) \Rightarrow Q)$. 
Hence, $(R \land (P \Rightarrow Q)) \Rightarrow ((R \Rightarrow P) \Rightarrow Q)$.
A: Don't start on the RHS, start on the LHS.
$$\begin{align}
& R \wedge (P\to Q) 
\\ \iff & \text{implication equivalence}
\\ & R\wedge (\neg P \vee Q) 
\\ \iff & \text{distribution}
\\ & (R\wedge \neg P)\vee (R\wedge Q) 
\\ \iff & \text{distribution}
\\ & [(R\wedge \neg P)\vee R]\wedge [(R\wedge \neg P)\vee Q] 
\\ \iff & \text{absorption}
\\ & R\wedge [(R\wedge \neg P)\vee Q] 
\\ \iff & \text{double negation then DeMorgan's}
\\ & R\wedge [\neg (\neg R\vee P)\vee Q] 
\\ \iff &\text{implication equivalence}
\\ & R\wedge [\neg (R\to P)\vee Q] 
\\ \iff &\text{implication equivalence}
\\ & R\wedge [(R\to P)\to Q] 
\end{align}$$
Now $R\wedge S \to S$ is a tautology, and we have the RHS as $S$, so we're done.
$$\therefore (R\wedge (P\to Q))\to ((R\to P)\to Q))$$

Alt. $R\wedge S \vDash S$ if you prefer.
