Prove $q \lor r \vdash (q \land s) \lor (s \to r)$ using natural deduction Am I going in the right direction? I'm stuck after this point and I don't really know what to do from here. I know I have to assume and introduce s but I'm unsure how.


*

*q v r premise

*q ʌ s assumption

*q ʌe 2,3 

*s ʌe 2,4

*r assumption 

 A: Fitch style natural deduction.  Unfortunately the excluded middle assumption on $s$ is not avoidable.
$$\begin{array} {rll}
(1) & q \lor r & \text{Given} \\
\\
(2) & \quad \quad q & \text{Premise} \\
(3) & \quad \quad s \lor \lnot s & \text{Excluded Middle on s} \\
(4) & \quad \quad \quad \quad s & \text{Premise} \\
    & \quad \quad \quad \quad \vdots & \text{Fill these in} \\
(5) & \quad \quad \quad \quad (q \land s) \lor (s \to r) &  \\
\\
(6) & \quad \quad \quad \quad \lnot s & \text{Premise} \\
    & \quad \quad \quad \quad \vdots & \text{Fill these in} \\
(7) & \quad \quad \quad \quad (q \land s) \lor (s \to r) &  \\
\\
(8) & \quad \quad (q \land s) \lor (s \to r) & \text{Or Elimination of 3, 4 to 5, 6 to 7} \\
\\
(9) & \quad \quad r & \text{Premise} \\
    & \quad \quad \vdots & \text{Fill in the dots} \\
(10) & \quad \quad (q \land s) \lor (s \to r) & \\
\\
(11) & (q \land s) \lor (s \to r) & \text{Or Elimination of 1, 2 to 8, 9 to 10} \\
\end{array}$$
That's the basic outline.  Let me know if you need hints filling in the dots.
A: So, you start with a disjunction.  Thus, you might manage to use disjunction elimination to solve this.  
If you assume r, hopefully, you can see how you can get to ((q$\land$s)$\lor$(s$\rightarrow$r)).
Now suppose that you assume 1. q.
Next assume 2. $\lnot$((q$\land$s)$\lor$(s$\rightarrow$r)).
That hopefully will enable you to produce $\lnot$(q$\land$s).
Once you have that, hopefully you can produce ($\lnot$q$\lor$$\lnot$s).
But, we have q still in effect.  So, hopefully you have some way that you can now produce $\lnot$s.
Hopefully then you can produce (s$\rightarrow$r) which makes getting to ((q$\land$s)$\lor$(s$\rightarrow$r) possible in one step.
Then we discharge 2. and somehow get to ((q$\land$s)$\lor$(s$\rightarrow$r)) with just 1. still under effect.
That hopefully makes using disjunction elimination with the premiss (q$\lor$r) possible. 
