course mth 151
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
014. Truth Tables
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q001. There are 8 questions in this set.
If each of the propositions p and q can be either true or false, what combinations of truth values are possible for the two propositions (e.g., one possibility is that p is false and q is true; list the other possibilities)?
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Your solution:
P Q
T T
T F
F T
F F
Confidence Assessment:
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Given Solution:
It is possible that p is true and q is true.
Another possibility is that p is true and q is false.
A third possibility is that p is false and q is true.
A fourth possibility is that p is false and q is false.
These possibilities can be listed as TT, TF, FT and FF, where it is understood that the first truth value is for p and the second for q.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q002. For each of the for possibilities TT, TF, FT and FF, what is the truth value of the compound statement p ^ q ?
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Your solution:
P Q P ^ Q
T T T
T F F
F T F
F F F
Confidence Assessment:
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Given Solution:
p ^ q means 'p and q', which is only true if both p and q are true.
In the case TT, p is true and q is true so p ^ q is true.
In the case TF, p is true and q is false so p ^ q is false.
In the case FT, p is false and q is true so p ^ q is false.
In the case FF, p is false and q is false so p ^ q is false.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q003. Write the results of the preceding problem in the form of a truth table.
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Your solution:
I already did answering the preceding problem.
Confidence Assessment:
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Given Solution:
The truth table must have headings for p, q and p ^ q. It must include a line for each of the possible combinations of truth values for p and q. The table is as follows:
p q p ^ q
T T T
T F F
F T F
F F F.
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Self-critique (if necessary):
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Question: **** `q004. For each of the possible combinations TT, TF, FT, FF, what is the truth value of the proposition p ^ ~q?
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Your solution:
P Q P ^ ~ Q
T T T F F T
T F T T T F
F T F F F T
F F F F T F
Confidence Assessment:
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Given Solution:
For TT we have p true, q true so ~q is false and p ^ ~q is false.
For TF we have p true, q false so ~q is true and p ^ ~q is true.
For FT we have p false, q true so ~q is false and p ^ ~q is false.
For FF we have p false, q false so ~q is true and p ^ ~q is false.
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Self-critique (if necessary):
After doing the table I don? know what row or column is the solution. I think its under the negation column.
Self-critique Rating:
Question: **** `q005. Give the results of the preceding question in the form of a truth table.
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Your solution:
I only know how to answer these using the table which is above.
Confidence Assessment:
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Given Solution:
The truth table will have to have headings for p, q, ~q and p ^ ~q. We therefore have the following:
p q ~q p^~q
T T F F
T F T T
F T F F
F F T F
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Self-critique (if necessary):
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Question: **** `q006. Give the truth table for the proposition p U q, where U stands for disjunction.
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Your solution:
P Q P V Q
T T T T T
T F T T F
F T F T T
F F F F F
Confidence Assessment:
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Given Solution:
p U q means 'p or q' and is true whenever at least one of the statements p, q is true. Therefore p U q is true in the cases TT, TF, FT, all of which have at least one 'true', and false in the case FF. The truth table therefore reads
p q p U q
T T T
T F T
F T T
F F F
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Self-critique (if necessary):
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Question: **** `q007. Reason out the truth values of the proposition ~(pU~q).
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Your solution:
~(PV~Q) = ~P ^ ~ Q
P Q ~P ^ ~Q
T T F F F
T F F F T
F T T F F
F F T T T
Confidence Assessment:
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Given Solution:
In the case TT p is true and q is true, so ~q is false. Thus p U ~q is true, since p is true. So ~(p U ~q) is false.
In the case TF p is true and q is false, so ~q is true. Thus p U ~q is true, since p is true (as is q). So ~(p U ~q) is false.
In the case FT p is false and q is true, so ~q is false. Thus p U ~q is false, since neither p nor ~q is true. So ~(p U ~q) is true.
In the case FF p is false and q is false, so ~q is true. Thus p U ~q is true, since ~q is true. So ~(p U ~q) is false.
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Self-critique (if necessary):
I though with ?nd?if there is a false then it makes it false.
The question is about an 'or' statement; however you can use deMorgan's laws, as you did (almost) correctly.
Note that
~(PV~Q) = ~P ^ ~ (~Q) = ~P ^ Q, not ~P ^ ~Q.
Self-critique Rating:
Question: **** `q008. Construct a truth table for the proposition of the preceding question.
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Your solution:
I already did.
&#Good responses. See my notes and let me know if you have questions. &#