Assignment 14

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course MTH 151

2:50 03/18/13

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 true

• p is false and q is false

• p is true and q is false

• p is false and q is true

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:

• T T - true

• T F - false

• F T - false

• F F - false

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):

Should I have went all out and type in

“ 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 Rating: OK

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

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Your solution:

• p q p^q

• TTT

• TFF

• FTF

• FFF

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?

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Your solution:

It would be the exact opposite as before

• TT - false

• TF - true

• FT - false

• FF - false

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

• TFT

• FTT

• FFF

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:

~(p U ~q)

• T T - false

• T F - false

• F T - true

• F F - false

confidence rating #$&*:

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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 pU~q ~(pU~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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&#Very good work. Let me know if you have questions. &#