Major Quiz

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20:21:07 `q001. There are 8 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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RESPONSE --> p T q F p F q F p T q T

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20:23:22 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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RESPONSE --> right

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20:24:14 `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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RESPONSE --> It would be TFFF

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20:24:26 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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RESPONSE --> right

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20:25:42 `q003. Write the results of the preceding problem in the form of a truth table.

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RESPONSE --> p q p^q T T T T F F F T F F F F

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20:25:52 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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RESPONSE --> right

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20:26:46 `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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RESPONSE --> It is FTFF

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20:26:57 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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RESPONSE --> RIGHT

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20:28:22 `q005. Give the results of the preceding question in the form of a truth table.

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RESPONSE --> p ~q p^~q T F F T T T F F F F T F

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20:28:32 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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RESPONSE --> OK

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20:30:13 `q006. Give the truth table for the proposition p U q, where U stands for disjunction.

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RESPONSE --> p q p V q T T T T F T F T T F F F

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20:30:34 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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RESPONSE --> RIGHT

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20:33:54 `q007. Reason out the truth values of the proposition ~(pU~q).

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RESPONSE --> FIRST MAKE TRUTH TABLE FOR P V ~Q p ~q p V ~q T F T T T T F F F F T T Then you apply the negation ~ T becomes F, T becomes F again and so on to get: FFTF

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20:34:33 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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RESPONSE --> right

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20:35:04 `q008. Construct a truth table for the proposition of the preceding question.

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RESPONSE --> I did that on the previous problem.

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20:35:42 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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RESPONSE --> do we have to write out the entire thing or can we use a shortcut method as I did?

It's safest to include the columns for p and q, but if you are very sure that you are using the standard order it's OK to leave those columns off.

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Œ¡ê~ûqú™¶çb¡Å¦Å©‘„Kî¤øž¢½OÁP¨ Student Name: assignment #015

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20:42:04 `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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RESPONSE --> p q p -> q T T T T F F F T T F F T

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20:42:21 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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RESPONSE --> RIGHT

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20:44:34 `q002. Reason out, then construct a truth table for the proposition ~p -> q.

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RESPONSE --> p q ~p ~p -> q T T F T T F F T F T T T F F T F

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20:45:37 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 F T since (T -> T) is T F F T T since (T -> F) is F

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RESPONSE --> OK

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20:50:40 `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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RESPONSE --> First I drew the truth table with all the headings, p, q, ~p, ~q, p ^ ~q, (~p -> ~q). Using the truth values for each column I ended up with (p ^ ~q ) V (~p -> ~q) having the values TTFT.

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20:52:12 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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RESPONSE --> I missed something in the problem here. The truth values were already assigned to the statements?

For this question, they were.

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20:56:15 `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ).

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RESPONSE --> This is what I did on the previous problem. If I missed something in the earlier problem, I aplologize. p q ~p ~q (p ^ ~q) (~p -> ~q) [(p^~q)V(~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

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20:57:31 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 T T T T F F T F 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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RESPONSE --> OK

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21:00:08 `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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RESPONSE --> The truth table is as follows: 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

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21:01:08 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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RESPONSE --> right

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21:08:00 `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r.

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RESPONSE --> I constructed another truth table. For the line TFT, (p ^ ~q) would be Tsince p and ~q are both T. Then, (p ^ ~q) -> r would be T -> T = T. Following the same logic for each: FFT is T FTF is T

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21:08:31 We would need column headings p, q, r, ~q, (p^~q) and (p^~q) -> r. The truth table would then read

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RESPONSE --> something must not have went through on the explanation.

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21:08:46 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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RESPONSE --> there it is.

That doesn't happen often, but there's an occasional editing error that splits the response like that.

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"

This looks good. Let me know if you have questions.