Assignment 15

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course MTH 151

3/13/14, 10:34pm

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.

015. Conditionals

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Question: `q001. There are 7 questions in this set.

The proposition p -> q is true unless p is true and q is false. Construct the truth table for this proposition.

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Your solution:

TTT, FFT, FTT, TFF

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A properly constructed truth table will look like this

p q p -> q

T T T

T F F

F T T

F F T

rather than this

TTT, FFT, FTT, TFF

However I can see what you mean, and your lines appear to be correct.

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confidence rating #$&*: 2

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Given Solution:

The proposition will be true in every case except the one where p is true and q is false, which is the TF case. The truth table therefore reads as follows:

p q p -> q

T T T

T F F

F T T

F F T

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Self-critique (if necessary):

After last week’s assignments with similar problems, I have studied the chapter more, and hope that I am understanding it better, but still feel like I am struggling a bit.

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Self-critique Rating: 3

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Question: `q002. Reason out, then construct a truth table for the proposition ~p -> q.

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Your solution:

FFTT, FTTT, TTFT, TFFT

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Without the column headings (e.g., p q ~p ~p -> q) it is impossible to know exactly what you mean. However if your headings are the same as those used in the solution, your results are OK.

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confidence rating #$&*: 2

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Given Solution:

This proposition will be false in the T -> F case where ~p is true and q is false. Since ~p is true, p must be false so this must be the FT case. The truth table will contain lines for p, q, ~p and ~p -> q. We therefore get

p q ~p ~p -> q

T T F T since (F -> T) is T

T F F T since (F -> F) is T

F T T T since (T -> T) is T

F F T T since (T -> F) is F

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Self-critique (if necessary):

OK

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Self-critique Rating: 3

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Question: `q003. Reason out the truth value of the proposition (p ^ ~q) U (~p -> ~q ) in the case FT (i.e., p false, q true).

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Your solution:

I really struggled to understand this problem, I think mainly because the symbols and parenthesis are throwing me off a little, and making it difficult for me to come up with a truth value.

confidence rating #$&*: 1

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Given Solution:

To evaluate the expression we must first evaluate p ^ ~q and ~p -> ~q.

p ^ ~q is evaluated by first determining the values of p and ~q. If p is false and q true, then ~q is false. Thus both p and ~q are false, and p ^ ~q is false.

~p -> ~q will be false if ~p is true and ~q is false; otherwise it will be true. In the FT case p is false to ~p is true, and q is true so ~q is false. Thus it is indeed the case the ~p -> ~q is false.

(p ^ ~q) U (~p -> ~q ) will be false if (p ^ ~q) and (~p -> ~q ) are both false, and will otherwise be true. In the case of the FT truth values we have seen that both (p ^ ~q) and (~p -> ~q ) are false, so that (p ^ ~q) U (~p -> ~q ) is false.

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Self-critique (if necessary):

ok

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&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

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Self-critique Rating: 3

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Question: `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ).

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Your solution:

Like the problem above, I struggled to understand this problem as well, mainly because of the symbols. I struggled to even come up with a guess for this problem.

confidence rating #$&*: 3

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Given Solution:

We will need headings for p, q, ~p, ~q, (p ^ ~q), (~p -> ~q ) and (p ^ ~q) U (~p -> ~q ). So we set up our truth table

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

T T F F F T T

T F F T T T T

F T T F F F F

F F T T F T T

To see the first line, where p and q are both T, we first see that ~p and ~q must both be false. (p ^ ~q) will therefore be false, since ~q is false; (~p -> ~q) is of the form F -> F and is therefore true. Since (~p -> ~q) is true, (p ^ ~q) U (~p -> ~q ) must be true.

To see the second line, where p is T and q is F, we for see that ~p will be F and ~q true. (p ^ ~q) will therefore be true, since both p and ~q are true; (~p -> ~q) is of the form F -> T and is therefore true. Since (p ^ ~q) and (~p -> ~q ) are both true, (p ^ ~q) U (~p -> ~q ) is certainly true.

To see the fourth line, where p is F and q is F, we for see that ~p will be T and ~q true. (p ^ ~q) will be false, since p is false; (~p -> ~q) is of the form T -> T and is therefore true. Since (~p -> ~q ) is true, (p ^ ~q) U (~p -> ~q ) is true.

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Self-critique (if necessary):

ok

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&#This also requires a self-critique.

&#

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Self-critique Rating: 3

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Question: `q005. If we have a compound sentence consisting of three statements, e.g., p, q and r, then what possible combinations of truth values can occur?

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Your solution:

With p & q, the truth values would be TT, TF, FF, FT. With all 3 combinations of p, q, and r, we would have TTT, FFF, TTF, FFT, TFT, FTF, FTT, TFF.

confidence rating #$&*: 2

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Given Solution:

A compound statement with two statements p and q has four possible combinations of truth values: TT, TF, FT, FF. Here we also have r, which can be either T or F. So we can append either T or F to each of the possible combinations for p and q.

If r is true then we have possible combinations TT T, TF T, FT T, FF T. If r is false we have TT F, TF F, FT F, FF F. This gives us 8 possible combinations: TTT, TFT, FTT, FFT, TTF, TFF, FTF, FFF.

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Self-critique (if necessary):

OK

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Self-critique Rating: 3

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Question: `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r.

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Your solution:

Really having trouble understanding this problem as well, for the same reasons. I’m not sure why the symbols and parenthesis are throwing me off so much.

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You really need to give me a more extensive and detailed self-critique.

However:

You want to evaluate (p ^ ~q) -> r

This statement is a conditional with antecedent p ^ ~q, and consequent r.

So to evaluate this you need columns for p ^ ~q, and for r.

If you set up the table as

p q r

T F T

F F T

F T F

you will have the r values, but you will still need the column for p ^ ~q. This expression should therefore go into the next-to-last column.

To evaluate the expression p ^ ~q you need a column for p and another for ~q. You will already have the column for p, but you will need a column for ~q, which should therefore preceded the column for p ^ ~q.

Thus you need column headings

p, q, r, ~q, (p^~q) and (p^~q) -> r

as indicated in the given solution.

You should probably review as well the earlier qa's and queries involving truth tables, where the logic of setting up truth tables is explained more fully than in this qa.

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confidence rating #$&*: 3

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Given Solution:

We would need column headings p, q, r, ~q, (p^~q) and (p^~q) -> r. The truth table would then read

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

T F T T T T

F F T T F T

F T F F F T

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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Question: `q007. Construct a truth table for the statement ~p -> q

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Your solution:

confidence rating #$&*:

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Self-critique Rating:

Going to re-read the chapter and study the chapter again to see if I can gain a better understanding of everything.

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Self-critique (if necessary):

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Self-critique rating:

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See if my notes help, and let me know what specific questions you have about my notes or the problems and solutions in this and in earlier queries and qa's.

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