Open qa 14

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course Mth 151

7/2 10

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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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 is true and q is false

P is true and q is true

P is false and q is false

confidence rating #$&*:3

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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):OK

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Self-critique Rating:OK

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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:

1. T ^ T = T

2. T ^ F = F

3. F ^ T =F

4. F ^ F= F

confidence rating #$&*:3

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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):OK

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Self-critique Rating:OK

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Question: `q003. Write the results of the preceding problem in the form of a truth table.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

P q p ^ q

T T F

T F F

F T F

F F F

confidence rating #$&*: 3

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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):OK

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Self-critique Rating:OK

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1. TT-TF- F

2. TF-TT-T

3. FT-FF-F

4. FF-FT-F

confidence rating #$&*:3

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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):OK

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Self-critique Rating:OK

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Question: `q005. Give the results of the preceding question in the form of a truth table.

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Your solution:

P q ~q p ^ ~q

T T F F

T F T T

F T F F

F F T F

confidence rating #$&*: 3

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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):OK

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Self-critique Rating:OK

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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 U q

T T T

T F T

F T T

F F F

confidence rating #$&*: 3

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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):OK

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Self-critique Rating:OK

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Question: `q007. Reason out the truth values of the proposition ~(pU~q).

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Your solution:

In TT if p is true and q is true, ~q is false, p U ~q is True, so ~ ( p U ~ q) is false

In TF if p is true and q is false, ~q is true, p U ~ q is True , so ~ ( p U ~ q) is False.

In FT if p is false and q is true, ~q is false, p U ~ q is False, so ~ ( p U ~ q) is true.

In FF if p is false and q is false, ~q is true, p U ~q is True, so ~ ( p U ~ q) is false.

confidence rating #$&*:3

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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):OK

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Self-critique Rating:OK

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Question: `q008. Construct a truth table for the proposition of the preceding question.

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Your solution:

P q ~q p U ~q ~(p U ~q)

T T F T F

T F T T F

F T F F T

F F T T F

confidence rating #$&*:3

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Given Solution:

We need headings for p, q, ~q, p U ~q and ~(p U ~q). Our truth table therefore read as follows:

p q ~q pU~q ~(pU~q)

T T F T F

T F T T F

F T F F T

F F T T F

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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