document.write( "Question 1190300: Topics In Contemporary Math
\n" ); document.write( " Modus Ponens and Modus Tollens\r
\n" ); document.write( "\n" ); document.write( "Another invalid argument form is the Fallacy of the Inclusive β€œor”, which has the
\n" ); document.write( "argument form
\n" ); document.write( "𝑝 𝑉 π‘ž
\n" ); document.write( "𝑝
\n" ); document.write( "∴ ~π‘ž
\n" ); document.write( "Create a truth table to prove that this argument form is invalid.
\n" ); document.write( "Translate each of the following into symbols, then determine whether or not the argument
\n" ); document.write( "is valid by providing the appropriate name for the argument form.
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Algebra.Com's Answer #821916 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Premise 1: P v Q
\n" ); document.write( "Premise 2: P
\n" ); document.write( "Conclusion: ~Q\r
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\n" ); document.write( "\n" ); document.write( "One way to form the truth table\n" ); document.write( "\n" ); document.write( "
Premise 1Premise 2Conclusion
PQP v QP~Q
TTTTF
TFTTT
FTTFF
FFFFT
Row 1, marked in red, shows all true premises lead to a false conclusion. This is sufficient to prove the argument is invalid.\r
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\n" ); document.write( "\n" ); document.write( "Here's an alternative way to form the truth table
\n" ); document.write( "What we do is conjunct the list of premises to form the antecedent, and this will lead to the conclusion.
\n" ); document.write( "(P v Q) & P is the antecedent while ~Q is the conclusion
\n" ); document.write( "This forms the conditional [ (P v Q) & P ] -> ~Q
\n" ); document.write( "If that is ever false, for any row, then we have proven the argument is invalid.
\n" ); document.write( "This is because we have true premises point to a false conclusion. \r
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\n" ); document.write( "\n" ); document.write( "This is what the truth table looks like using this alternative method\n" ); document.write( "\n" ); document.write( "
PQP v Q(P v Q) & P~Q[ (P v Q) & P ] -> ~Q
TTTTFF
TFTTTT
FTTFFT
FFFFTT
We have \"F\" at the very end of the first row (in red) to show that [ (P v Q) & P ] -> ~Q is false when P = T and Q = T
\n" ); document.write( "This confirms what the other table is showing (also in row 1).\r
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\n" ); document.write( "\n" ); document.write( "Some side notes:
  • P v Q is false if both P and Q are false, otherwise it's true.
  • P & Q is true if both P and Q are true, otherwise it's false.
  • P -> Q is false if P = T and Q = F, otherwise it's true.

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