Assignment 17

course Mth151

xž¼¤£µž’¸ªŸhèÛn¡ÿŒ{íâŠàñStudent Name:

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

assignment #017

.................................................

......!!!!!!!!...................................

08:41:33

`q001. There are 9 questions in this set.

Explain why [ (p -> q) ^ (q -> r) ^ p] -> r must be true for every set of truth values for which r is true.

......!!!!!!!!...................................

RESPONSE -->

because r is on the right side of the -> which means that no matter what the truth value of the left side, it will always read if ___ then True... this means the statment is always true

.................................................

......!!!!!!!!...................................

08:41:36

[ (p -> q) ^ (q -> r) ^ p] -> r must be true if r is true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. Therefore the truth values TTT, TFT, FTT, FFT (i.e., all the truth values that have r true) all make the statement true.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:42:23

`q002. At this point we know that the truth values TTT, TFT, FTT, FFT all make the argument [ (p -> q) ^ (q -> r) ^ p] -> r true. What about the truth values TTF?

......!!!!!!!!...................................

RESPONSE -->

The statement would be false because it would be true on the left side of the statement and always false on the right... if _____ then false.....

.................................................

......!!!!!!!!...................................

08:42:26

It would be possible to evaluate every one of the statements p -> q, q -> r, etc. for their truth values, given truth values TTF. However we can shortcut the process.

We see that [ (p -> q) ^ (q -> r) ^ p] is a compound statement with conjunction ^. This means that [ (p -> q) ^ (q -> r) ^ p] will be false if any one of the three compound statements p -> q, q -> r, p is false.

For TTF we see that one of these statements is false, so that [ (p -> q) ^ (q -> r) ^ p] is false. This therefore makes the statement [ (p -> q) ^ (q -> r) ^ p] -> r true.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:43:06

`q003. The preceding statement said that for the TTF case [ (p -> q) ^ (q -> r) ^ p] was false but did not provide an explanation of this statement. Which of the statements is false for the truth values TTF, and what does this tell us about the truth of the statement [ (p -> q) ^ (q -> r) ^ p] -> r?

......!!!!!!!!...................................

RESPONSE -->

already answered...oops

.................................................

......!!!!!!!!...................................

08:43:16

p and q are both true, so p -> q and p are true. The only candidate for a false statement among the three statements is q -> r.

So we evaluate q -> r for truth values TTF. Since q is T and r is F, we see that q -> r must be F.

This makes [ (p -> q) ^ (q -> r) ^ p] false. Therefore [ (p -> q) ^ (q -> r) ^ p] -> r must be true, since it can only be false and if [ (p -> q) ^ (q -> r) ^ p] is true.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:47:35

`q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF.

......!!!!!!!!...................................

RESPONSE -->

for TFF for FTF FFF

p->q is false p->q is true p->q is true

and and and

q->r is true q->r is false q->r is true

and and and

p is true... p is false p is false

F and T and T T & F & F F & F & F

false statement false statement false stmnt

.................................................

......!!!!!!!!...................................

08:47:39

In the case TFF, p is true and q is false so p -> q is false, which makes [ (p -> q) ^ (q -> r) ^ p] false.

In the case FTF, p is false, making [ (p -> q) ^ (q -> r) ^ p] false.

In the case FFF, p is again false, making [ (p -> q) ^ (q -> r) ^ p] false.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:48:01

`q005. We have seen that for TFF, FTF and FFF the statement [ (p -> q) ^ (q -> r) ^ p] is false. How does this help us establish that [ (p -> q) ^ (q -> r) ^ p] -> r is always true?

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:48:06

The three given truth values, plus the TTF we examined earlier, are all the possibilities where r is false. We see that in the cases where r is false, [ (p -> q) ^ (q -> r) ^ p] is always false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true any time r is false.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:49:15

`q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true.

......!!!!!!!!...................................

RESPONSE -->

because when r is true, then the statement is always true (if ___ then r(true))

.................................................

......!!!!!!!!...................................

08:49:19

We just finished showing that if r is false, [ (p -> q) ^ (q -> r) ^ p] is false so [ (p -> q) ^ (q -> r) ^ p] -> r is true. As seen earlier the statement must also be true whenever r is true. So it's always true.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:50:42

`q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid.

......!!!!!!!!...................................

RESPONSE -->

p->q

q->r

------

the statement when symbolized is the same....

.................................................

......!!!!!!!!...................................

08:50:46

That argument is symbolized by the statement [ (p -> q) ^ (q -> r) ^ p] -> r. The statement is always true. There is never a case where the statement is false. Therefore the argument is valid.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:52:21

`q008. Explain how the conclusion of the last example, that [ (p -> q) ^ (q -> r) ^ p] -> r is always a true statement, shows that the following argument is valid: 'If it snows, the roads are slippery. If the roads are slippery they'll be safer to drive on. It just snowed. Therefore the roads are safer to drive on.' Hint: First symbolize the present argument.

......!!!!!!!!...................................

RESPONSE -->

p->q

q->r

---------

again, the argument is symbolized the same way.... r is true...if is snows, the roads are safer to drive on (p->r)

.................................................

......!!!!!!!!...................................

08:52:35

This argument can be symbolized by letting p stand for 'it snows', q for 'the roads are slippery', r for 'the roads are safer to drive on'. Then

'If it snows, the roads are slippery' is symbolized by p -> q.

'If the roads are slippery they'll be safer to drive on' is symbolized by q -> r.

'It just snowed' is symbolized by p.

'The roads are safer to drive on' is symbolized by r.

The argument the says that IF [ p -> q, AND q -> r, AND p ] are all true, THEN r is true.

In symbolic form this is [ (p -> q) ^ (q -> r) ^ p] -> r. This is the same statement as before, which we have shown to be always true. Therefore the argument is valid.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

08:53:44

`q009. Symbolize the following argument and show that it is valid: 'If it doesn't rain there is a picnic. There is no picnic. Therefore it rained.'

......!!!!!!!!...................................

RESPONSE -->

~p->q

-----------

or (~p->q)^~p->p

.................................................

......!!!!!!!!...................................

08:53:54

We could let p stand for 'it rained', q for 'there is a picnic'. The first statement is 'If it doesn't rain there is a picnic', which is symbolized by ~p -> q. The second statement, 'There is no picnic', is symbolized by ~q. The conclusion, 'it rained', is symbolized by p.

The argument therefore says IF [ (~p -> q) AND ~q ], THEN p. This is symbolized by [ (~p -> q) ^ ~q ] -> p.

We set up a truth table for this argument:

p q ~p ~q ~p -> q (~p -> q) ^ ~q [ (~p -> q) ^ ~q ] -> p

T T F F T F T

T F F T T T T

F T T F T F T

F F T T F F T

......!!!!!!!!...................................

RESPONSE -->

................................................."

Assignment 17

Your work on this assignment is good. Let me know if you have questions.