Asgt 15 QA

course MTH 151

10:35 PM 10/19/09

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

.

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 6 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:

p q p -> q

T T T

T F F

F T T

F F T

confidence rating:3

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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):

Ok

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Self-critique Rating:

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

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Your solution:

If negation of p then q. This means p would have to be false and q would have to be false so that you would end up with a TF statement. So all values true except the FF.

p q -p -> q

T T F T T

T F F T F

F T T T T

F F T F F

confidence rating:3

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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:

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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:

The conjunction would be F^F so it would be False

The conditional would be T -> F so it would also be false

The disjunction would be F U F so it would also be false

The entire statement is False

confidence rating:3

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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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Self-critique Rating:

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

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Your solution:

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

T T T F F T F T F

T F T T T T F T T

F T F F F F T F F

F F F F T T T T T

So the outcomes are TTFT

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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Self-critique Rating:

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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:

p q r

T T T

T T F

T F T

T F F

F T T

F T F

F F T

F F F

Interesting how that pattern works.... for p its 4T and 4F, for q its 2T 2F 2T 2F, for r it alternates TF

Right. There are a number of equivalent ways of defining the pattern. One is 'in each line use as few F's as possible and keep them as far to the right as possible, without repeating any of the other lines'. Another is 'start with the column T F. At every step the new table repeats the existing columns twice, appending a column of T's in front of the first repetition and a column of F's in front of the second'.

confidence rating:3

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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:

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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:

TFT:

we have T ^ T -> T

This translates to T -> T which is True

FFT:

F ^ T -> T

F > T is True

FTF:

F ^ F > F

F > F is True

None of the above situations gave a TF conditional so they were all true

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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&#Good work. See my notes and let me know if you have questions. &#