Crystal Call

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course mth 151

03/13/2014assignment 14

Question: `q001. There are 10 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 q is false

P is true q is true

P is false q is false

P is false q is true

confidence rating #$&*:

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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):

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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:

P is true, q is true = true

P is true, q is false = false

P is false, q is true = false

P is false, q is false = false

confidence rating #$&*:

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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):

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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

T T T

T F F

F T F

F F T

confidence rating #$&*:

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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 true, q true so ~q is false and p ^ ~q is false.

p true, q false so ~q is true and p ^ ~q is true.

p false, q true so ~q is false and p ^ ~q is false.

p false, q false so ~q is true and p ^ ~q is false.

confidence rating #$&*:

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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):

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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 #$&*:

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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 U Q

T T T

T F T

F T T

F F F

confidence rating #$&*:

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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:

Work with in parenthesis first

~ ( p U ~q)

~ ( t U f )

~ (t)

false

I got the same answer as you did for all of the given solution, but I worked it more like an algebraic equations where a negative and a positive multiplied by each other makes it negative…or in this case, false. Can it work that way or was this time just a ‘lucky shot’

@&

If you negate a true statement you get a false and if you negate a false statement you get a true. So negation works in a manner that is completely analogous to multiplying by -1, in that the truth value changes just as the sign of the number changes.

However the analogy doesn't completely work when you try to apply a distributive law, where deMorgan's laws apply.

*@

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):

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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 #$&*:

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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):

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Question: `q009. Construct a truth table for the statement (p ^ ~q).

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Your solution:

P Q ~Q (p ^ ~q)

T T F F

T F T F

F T F F

F F T T

confidence rating #$&*:

@&

p ^ ~q is a conjunction, which is true when and only when both of the statements are true.

The statements connected by the conjunction are p and ~q. So the statement is true when both p and ~q are both true.

In your last line p is false, so p ^ ~q is false.

There is a line in your table where p is true and ~q is true. That is the line for which the statement p ^ ~q is true.

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Question: `q010. Construct a truth table for the statement q U (p ^ ~q).

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Your solution:

P Q ~Q (p ^ ~q)

T T F T

T F T T

F T F T

F F T F

Confidence Rating:

These past two problems, I am not sure if I did it correctly. So my confidence is just so-so

@&

Your table is not consistent with the table of the preceding problem. However that table had at least one error. So it is possible that this table is actually correct. However you will need to determine whether this is indeed the case.

This statement is a disjunction, which is true as long as the two statements are not both false.

The two statement are q and p ^ ~q.

If you correct the preceding table for p ^ ~q you will have the truth values for that statement. Your table includes the values for q. Put the two together and be sure your table is consistent with the corrected table from the preceding, then see whether the table you give here is accurate. If not, revise it.

I recommend that you sent me copies of these two questions with your revised solution. If you're confident you have it, it isn't necessary for you to send that, but otherwise consider doing so.

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Self-critique Rating:

&#Your work looks good. See my notes. Let me know if you have any questions. &#