Question 1190300: Topics In Contemporary Math
Modus Ponens and Modus Tollens
Another invalid argument form is the Fallacy of the Inclusive βorβ, which has the
argument form
π π π
π
β΄ ~π
Create a truth table to prove that this argument form is invalid.
Translate each of the following into symbols, then determine whether or not the argument
is valid by providing the appropriate name for the argument form.
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Premise 1: P v Q
Premise 2: P
Conclusion: ~Q
One way to form the truth table
| | Premise 1 | Premise 2 | Conclusion | | P | Q | P v Q | P | ~Q | | T | T | T | T | F | | T | F | T | T | T | | F | T | T | F | F | | F | F | F | F | T |
Row 1, marked in red, shows all true premises lead to a false conclusion. This is sufficient to prove the argument is invalid.
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Here's an alternative way to form the truth table
What we do is conjunct the list of premises to form the antecedent, and this will lead to the conclusion.
(P v Q) & P is the antecedent while ~Q is the conclusion
This forms the conditional [ (P v Q) & P ] -> ~Q
If that is ever false, for any row, then we have proven the argument is invalid.
This is because we have true premises point to a false conclusion.
This is what the truth table looks like using this alternative method
| P | Q | P v Q | (P v Q) & P | ~Q | [ (P v Q) & P ] -> ~Q | | T | T | T | T | F | F | | T | F | T | T | T | T | | F | T | T | F | F | T | | F | F | F | F | T | T |
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
This confirms what the other table is showing (also in row 1).
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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