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MODULE TWO PROBLEM SET
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Alfredo J. Romero C.
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Directions: Type your solutions into this document and be sure to show all steps for arriving at your solution. Just giving a final number may not receive full credit.
PROBLEM 1
Part 1. Indicate whether the argument is valid or invalid. For valid arguments, prove that the argument is valid using a truth table. For invalid arguments, give truth values for the variables showing that the argument is not valid.
(1)
(bVa)
.. (p V q) > r
Given argument is:
sasiuad EF(OVd) .
. :. (P V Q) - - Conclusion
num IP|Q l(OVd)|a l(PV Q)|Premise(P A Q->R|Conclusion(PV Q)->R 1 T T T 2 T F 3 F T F T T T 4 F F T T F 5 2 T F T T T 6 T F F T T F 7 F 7 F F T T 8 F F F T T T for valid or invalid argument we have to check critical row.
: critical row @ rows with this symbol onthe right side. - Valid Arguments - if all critical row of premises has true value of all conclusion, then it is valid argument. : Invalid Arguments - if all critical row of premises has false value of all conclusion, then it is invalid argument. So, critical rows are 1,3,4,5,6,7,8 but values of 4 and 6 has false conclusion. so, the argument is invalid.
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Part 2. Converse and inverse errors are typical forms of invalid argu- ments. Prove that each argument is invalid by giving truth values for the variables showing that the argument is invalid. You may find it eas- ier to find the truth values by constructing a truth table.
a) Converse error
The truth for p to q is as follow:
num 1 2 T 3 4 F
T F T T
Since it is given than P to Q is true, pick the rows where the value of P to Q is true. This result in the following table:
num P
>Q T T T
3
Also, given that Q is true, pick the rows where the value of Q is true. this result in the following table:
num IPQP -Q T T
As per the given argument, the value of P should always come out to be true if the value of P to Q and Q is true. Since there exists a false assignment for p in the above table, the given argument is inval