course mth 151 ?xE????????s???Student Name:
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15:54:07 `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 t t t f f t f f p->q t f t t
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15:54:12 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 --> ok
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15:55:15 `q002. Reason out, then construct a truth table for the proposition ~p -> q.
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RESPONSE --> p q t t t f f t f f ~p->q f f t f
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15:55:31 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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15:56:34 `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 --> f
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15:56:49 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 --> f stood for false
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15:58:17 `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ).
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RESPONSE --> p q t t t f f t f f p ^ ~q f t f f ~p->q f t f f
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15:58:30 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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16:00:13 `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 --> ttt tft tff fff ftt ftf 6
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16:00:25 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 --> i missed two
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16:02:44 `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r.
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RESPONSE -->
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16:03:22 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 --> i hit enter too many times
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16:03:30 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 --> ok
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