course Mth 151 ????w????{?
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20:59:11 `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 --> confidence assessment:
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21:00:32 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 --> if p and q are true then p->q are true but if p is ture and q is false then p->q are false self critique assessment: 3
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21:05:11 `q002. Reason out, then construct a truth table for the proposition ~p -> q.
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RESPONSE --> if p is false then q is true: p q ~p ~q ~p->q T T F F F confidence assessment: 2
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21:07:52 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 --> becuase ~p is false then ~p->q is true self critique assessment: 3
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21:13:46 `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 --> [p and ~q] or [~p->~q] p is true and ~q is false or ~p is false if ~q is false confidence assessment: 3
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21:32:55 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 --> because p is false, ~p is true so ~p->~q is false self critique assessment: 3
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21:42:56 `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ).
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RESPONSE --> p q ~p ~q [p^ ~q] [~p->~q] T T F F F T T F F T T F F T T F T F F F T T T F confidence assessment: 3
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21:49:00 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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RESPONSE --> I need to analize the begining if the table and follow to come up with the answers as I head to the end self critique assessment: 3
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22:01:28 `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 --> 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 confidence assessment: 3
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22:03:46 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 --> we can add a value to r without 'bothering the the values on p q self critique assessment: 3
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22:07:42 `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r.
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RESPONSE --> [p^~q] -> r; meaning if pand ~q are true so is r; this statement will be false confidence assessment: 3
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22:09:05 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 --> false self critique assessment: 3