Question 1157893
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The conclusion is H, so the opposite of this is ~H


We'll use a proof by contradiction
The idea is to assume ~H is the case, and show that it leads to a contradiction. This will then mean the opposite of ~H, which is H, must be the true case instead.


If ~H is the case, then we can use modus tollens on premise 2 to get ~(D v G) which turns into ~D & ~G after using De Morgan's Law.


After using simplification, ~D & ~G turns into ~G. Use ~G with E v G, and you'll get E. Use the disjunctive syllogism rule here.


Now that we have E, use the addition rule to get E v F. Next up, apply modus ponens on premise 1 to get C & D which can be simplified to D.


Now go back to ~D & ~G. We can also simplify this to ~D. But this contradicts the D we got earlier in the last paragraph. This concludes the proof by contradiction and confirms that the argument as presented, with the conclusion H, is a valid argument. 


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Here is a derivation table which may be one format your teacher wants you to use
<table border = "1" cellpadding = "5">
<tr><td colspan=2>Number</td><td>Statement</td><td>Lines Used</td><td>Reason</td></tr>
<tr><td>1</td><td></td><td>(E v F) -> (C & D)</td><td></td><td></td></tr>
<tr><td>2</td><td></td><td>(D v G) -> H</td><td></td><td></td></tr>
<tr><td>3</td><td></td><td>E v G</td><td></td><td></td></tr>
<tr><td colspan=2>Conclusion</td><td>H</td><td></td><td></td></tr>
<tr><td></td><td>4</td><td>~H</td><td></td><td>Assumption for Indirect Proof</td></tr>
<tr><td></td><td>5</td><td>~(D v G)</td><td>2,4</td><td>Modus Tollens</td></tr>
<tr><td></td><td>6</td><td>~D & ~G</td><td>5</td><td>De Morgan’s Law</td></tr>
<tr><td></td><td>7</td><td>~D</td><td>6</td><td>Simplification</td></tr>
<tr><td></td><td>8</td><td>~G</td><td>6</td><td>Simplification</td></tr>
<tr><td></td><td>9</td><td>E</td><td>3,8</td><td>Disjunctive Syllogism</td></tr>
<tr><td></td><td>10</td><td>E v F</td><td>9</td><td>Addition</td></tr>
<tr><td></td><td>11</td><td>C & D</td><td>1,10</td><td>Modus Ponens</td></tr>
<tr><td></td><td>12</td><td>D</td><td>11</td><td>Simplification</td></tr>
<tr><td></td><td>13</td><td>D & ~D</td><td>7,12</td><td>Conjunction</td></tr>
<tr><td>14</td><td></td><td>H</td><td>4-13</td><td>Indirect Proof</td></tr>
</table>



Note how lines 7 and 12 contradict each other. 
Line 14 is the opposite of line 4.

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