#$&* course MTH 151 3/13/14, 10:34pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The proposition will be true in every case except the one where p is true and q is false, which is the TF case. The truth table therefore reads as follows: p q p -> q T T T T F F F T T F F T &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After last week’s assignments with similar problems, I have studied the chapter more, and hope that I am understanding it better, but still feel like I am struggling a bit. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. Reason out, then construct a truth table for the proposition ~p -> q. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: FFTT, FTTT, TTFT, TFFT
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Given Solution: This proposition will be false in the T -> F case where ~p is true and q is false. Since ~p is true, p must be false so this must be the FT case. The truth table will contain lines for p, q, ~p and ~p -> q. We therefore get p q ~p ~p -> q T T F T since (F -> T) is T T F F T since (F -> F) is T F T T T since (T -> T) is T F F T T since (T -> F) is F &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q003. Reason out the truth value of the proposition (p ^ ~q) U (~p -> ~q ) in the case FT (i.e., p false, q true). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I really struggled to understand this problem, I think mainly because the symbols and parenthesis are throwing me off a little, and making it difficult for me to come up with a truth value. confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q007. Construct a truth table for the statement ~p -> q YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: