Question 1190424
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T = true
F = false


P = Adora is a princess
H = she (Adora) fights the horde


~P = Adora is not a princess
~H = she does not fight the horde


Premise 1: P -> H
Premise 2: ~H
Conclusion: ~P


The arrow notation P -> Q means "If P, then Q".


Here's what the truth table looks like<table border = "1" cellpadding = "5"><tr><td></td><td></td><td>Premise 1</td><td>Premise 2</td><td>Conclusion</td></tr><tr><td>P</td><td>H</td><td>P -> H</td><td>~H</td><td>~P</td></tr><tr><td>T</td><td>T</td><td>T</td><td>F</td><td>F</td></tr><tr><td>T</td><td>F</td><td>F</td><td>T</td><td>F</td></tr><tr><td>F</td><td>T</td><td>T</td><td>F</td><td>T</td></tr><tr><td>F</td><td>F</td><td>T</td><td>T</td><td>T</td></tr></table>
P -> H is only true when P is true while H is false. Otherwise, the conditional statement is true.
~H will flip everything in the H column. 
~P flips everything in the P column.


As the table shows, we do not have any rows with all true premises lead to a false conclusion. 
This means we are not able to prove the argument is invalid.
Therefore, this argument is valid.


The argument is valid by the modus tollens rule.


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Here's another way to form the truth table and to show we have a valid argument<table border = "1" cellpadding = "5"><tr><td>P</td><td>H</td><td>P -> H</td><td>~H</td><td>(P -> H) & ~H</td><td>[ (P -> H) & ~H ] -> ~P</td></tr><tr><td>T</td><td>T</td><td>T</td><td>F</td><td>F</td><td>T</td></tr><tr><td>T</td><td>F</td><td>F</td><td>T</td><td>F</td><td>T</td></tr><tr><td>F</td><td>T</td><td>T</td><td>F</td><td>F</td><td>T</td></tr><tr><td>F</td><td>F</td><td>T</td><td>T</td><td>T</td><td>T</td></tr></table>The last column is all T, so there are no cases in which the argument is invalid.
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