Question 1208944
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Here's one way to do the derivation. Interestingly, premise #2 is never used so I don't know if I might have overlooked something or perhaps your teacher made a typo somewhere.
I'm using an arrow in place of a horseshoe. 
Also, I'm using an ampersand (&) in place of a dot. 
<table border = "1" cellpadding = "5"><tr><td>Number</td><td>Statement</td><td>Line(s) Used</td><td>Reason</td></tr><tr><td>1</td><td>~(Q --> ~R)</td><td></td><td></td></tr><tr><td>2</td><td>~(~P & Q)</td><td></td><td></td></tr><tr><td>:.</td><td>~R --> (P & Q)</td><td></td><td></td></tr><tr><td>3</td><td>~(~Q v ~R)</td><td>1</td><td>Material Implication</td></tr><tr><td>4</td><td>~(~Q) & ~(~R)</td><td>3</td><td>De Morgan’s Law</td></tr><tr><td>5</td><td>Q & R</td><td>4</td><td>Double Negation</td></tr><tr><td>6</td><td>R & Q</td><td>5</td><td>Commutation</td></tr><tr><td>7</td><td>R</td><td>6</td><td>Simplification</td></tr><tr><td>8</td><td>R v (P & Q)</td><td>7</td><td>Addition</td></tr><tr><td>9</td><td>~R --> (P & Q)</td><td>8</td><td>Material Implication</td></tr></table>
Here's a list of <a href="https://www.algebra.com/algebra/homework/Conjunction/logic-rules-of-inference-and-replacement.lesson">rules of inference and replacement</a>
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