course Mth 151
I left the original answers and italizied my new answers.
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
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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.
017. Evaluating Arguments
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `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.
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Your solution:
because it is a tautology? I realize I said this because all ->r is True when I did the following table:
p q r p->q q->r ^ ^p ->r
T T T T T T T T
T T F T F F F T
T F T F T F F T
T F F F T F F T
F T T T T T F T
F T F T F F F T
F F T T T T F T
F F F T T T F T
So what you are saying is the reason that [(p -> q) ^ (q -> r) ^ p] -> r is true is just because all of r is true and if it were all false then the whole thing is false? My confusion stems from the truth values such as TTT, TFT, FTT, FFT but I now see that from my table that TTT is the first row of p,q,r; TFT is the 3rd row of p,q,r; FTT is 5th row of p,q,r; and FFT is the 7th row of p,q,r. Am I understanding this correctly?
The given solution doesn't address what happens if r is false. However if the consequent of a conditional is true, the whether the antecedent is true or false, the conditional must be true.
(It's not relevant to this question, but if the consequent is false then, depending on whether the antecedent is true or false, the conditional is respectively false or true.)
Confidence Assessment:
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Given Solution:
[ (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.
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Self-critique (if necessary):
I still don’t quite grasp this. Is this the same thing as the table?
See above I believe I am getting it.
Self-critique Rating:
Question: **** `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?
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Your solution:
I think it is the way it is set up that confuses me. If I see it in a table it would be clearer.
p q r p->q q->r ^ ^p ->r
T T F T F F F T
It looks like it also makes it true.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
This is just what the book said when I was trying to answer it. It said if at least one premise in a conjunction of several premises is false then the entire conjunction is false.
O.K is this saying that [(p -> q) ^ (q -> r) ^ p] is false but with the conditional of r it makes it true?
[(p -> q) ^ (q -> r) ^ p] is the antecedent of the conditional and r the consequent.
If r is true then the conditional is true.
Self-critique Rating:
Question: **** `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?
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Your solution:
p q r p->q q->r [(p->q)^(q->r)^p] ->r
T T T T T T T
T T F T F T F
T F T F T T T
T F F F T T F
F T T T T T T
F T F T F T F
F F T T T T T
F F F T T T F
because it is a tautology.
It looks like I did my table wrong. So going by my table that is clearer I would say it starts at q->r.
The first five columns of your table are fine. I can't tell what you did in the last two, but they don't correspond to the expressions you've listed as headings.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
Explain to me about finding truth in these sets such as TTF. I can’t find it in the book nor did the lady on the video say anything about them. If it is in the book can you tell me what page? When I figure it out I will finish this assignment and resubmit it.
Self-critique Rating:
Question: **** `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF.
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Your solution:
p q r p->q q->r ^ ^p ->r
T F F F T F F T
F T F T F F F T
F F F T T T F T
With TFF it starts at p->q, FTF starts at q->r, and FFF starts at ^p for being false.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
So if p is false at the starting point that is where it makes it false but if p is true the next false would be in the conditional?
Self-critique Rating:
Question: **** `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?
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Your solution:
This is where I am not sure. I can see where the falses are but to know that [ (p -> q) ^ (q -> r) ^ p] -> r is always true I don’t know how to explain.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true.
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Your solution:
[ (p -> q) ^ (q -> r) ^ p]is always false but when you add the conditional r it makes it true because it is only false when [ (p -> q) ^ (q -> r) ^ p] is true and r is false.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid.
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Your solution:
We know that if it rains the grass gets wet and wet grass smells so if it rains you can smell wet grass.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
I get confused whether to use symbols or words.
Self-critique Rating:
Question: **** `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.
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Your solution:
p = it snows
q= the roads are slippery
r = safe to drive
[(p->q)^(q->r)^p]->r When you look at it you think this can’t be valid but according to the rules it is. It is getting clearer.
Confidence Assessment:
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Given Solution:
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.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `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.'
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Your solution:
p=rain
q=picnic
[(~p->q)^(~q->p)]
Confidence Assessment:
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Given Solution:
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
I left my conditional p out at the end (my conclusion) because I put it with my q why is this wrong?
When you insert new comments, mark them with &&&& so I can easily identify them. Also include any notes I've inserted. I've found and responded to some statements I don't remember from before, but I'm sure I haven't found them all. Add the marks to anything I've missed, add more questions if you have them, and resubmit.