#$&* course mth 151 assignment 153/14/214 Question: `q001. There are 7 questions in this set.
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Given Solution: The proposition will be true in every case except the one where p is true and q is false, which is the TF case. The truth table therefore reads as follows: p q p -> q T T T T F F F T T F F T &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Reason out, then construct a truth table for the proposition ~p -> q. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This will be false in the t f case where ~p is true and q is false. Since ~p is true, p must be false so this must be the FT case. The truth table will contain lines for p, q, ~p and ~p -> q p q ~p ~p -> q T T F T since (F -> T) is T T F F T since (F -> F) is T F T T T since (T -> T) is T F F T T since (T -> F) is F confidence rating #$&*: my confidence in these problems is very minimal. I am reading through the given solutions as well as looking through the chapter, but I am still not completely sure how these work and I don’t even know what questions to ask. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: 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. confidence rating #$&*: I am sorry but I am trying to figure these out and still get my assignments in on time…so I had to copy and paste ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To evaluate the expression we must first evaluate p ^ ~q and ~p -> ~q. p ^ ~q is evaluated by first determining the values of p and ~q. If p is false and q true, then ~q is false. Thus both p and ~q are false, and p ^ ~q is false. ~p -> ~q will be false if ~p is true and ~q is false; otherwise it will be true. In the FT case p is false to ~p is true, and q is true so ~q is false. Thus it is indeed the case the ~p -> ~q is false. (p ^ ~q) U (~p -> ~q ) will be false if (p ^ ~q) and (~p -> ~q ) are both false, and will otherwise be true. In the case of the FT truth values we have seen that both (p ^ ~q) and (~p -> ~q ) are false, so that (p ^ ~q) U (~p -> ~q ) is false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P q ~p ~q (p ^ ~q) (~p -> ~q ) (p ^ ~q) U (~p -> ~q ) T T F F F T T T F F T T T T F T T F F F F F F T T F T T confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A compound statement with two statements p and q has four possible combinations of truth values: TT, TF, FT, FF. Here we also have r, which can be either T or F. So we can append either T or F to each of the possible combinations for p and q. If r is true then we have possible combinations TT T, TF T, FT T, FF T. If r is false we have TT F, TF F, FT F, FF F. This gives us 8 possible combinations: TTT, TFT, FTT, FFT, TTF, TFF, FTF, FFF. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P q r ~q (p^~q) (p^~q) -> r T F T T T T F F T T F T F T F F F T confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Construct a truth table for the statement ~p -> q P q ~p ~p --> q F T T t T T F t F F T f T T F f YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course mth 151 Assignment 1603/14/2014 Question: `q001. There are 7 questions in this set.
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Given Solution: This argument certainly seems valid. We say what will happen if rains, and what will happen is that happens. Then we say that it rains, so the whole chain of happenings, rained then wet grass then smell, should follow. STUDENT QUESTION ??? What if the grass does not smell or there’s a person unable to smell? INSTRUCTOR RESPONSE None of that affects the logical validity of the argument. If the premises are accepted, then the conclusion is valid. Any argument about whether the premises are valid is not relevant to the logical validity of the argument. Of course if the premises aren't so then it's possible the conclusion isn't so either, but that doesn't prevent the argument from being logically valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Is the following argument valid: 'If it snows, the roads will be slippery. If the roads are slippery they'll be safer to drive on. Yesterday it snowed. Therefore yesterday the roads were safer to drive on.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would say according to the statements only, then yes the arguments are valid, though slippery roads are not safe to drive on. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The validity of an argument has nothing to do with whether the statements in that argument are true or not. All we are allowed to do is assume that the statements are indeed true, and see if the conclusions of the argument therefore hold. In this case, we might well question the statement 'if the roads are slippery they'll be safer to drive on', which certainly seems untrue. However that has nothing to do with the validity of the argument itself. We can later choose to reject the conclusion because it is based on a faulty assumption, but we cannot say that the argument is invalid because of a faulty assumption. This argument tells us that something will happen if it snows, and then tells us what we can conclude from that. It then tells us that it snows, and everything follows logically along a transitive chain, starting from from the first thing. STUDENT COMMENT: so it does not matter that the roads are not safer when they're slippery, what matters is that the statement said they are when snows and snowed yesterday therefore the roads were safer yesterday INSTRUCTOR RESPONSE: Right. The statements don't have to be true for the argument to be valid. Of course, if the statements aren't true then even though the argument is valid the conclusion might not be true. The old saying is 'garbage in, garbage out'. If you put 'garbage' (i.e., false statements) into a logical argument, that argument can indeed result in 'garbage' (i.e., a false statement as the logical conclusion). STUDENT COMMENT: According to the statement it is true, but I might question this about driving on slippery roads. INSTRUCTOR RESPONSE: That assumption is deliberately absurd, to help make a clear distinction between correct assumptions and correct logic. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Is the following argument valid: 'Today it will rain or it will snow. Today it didn't rain. Therefore today it snowed.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is a valid statement being that one OR the other will happen. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we accept the fact that it will do one thing or another, then at least one of those things must happen. If it is known that if one of those things fails to happen, then, the other must. Therefore this argument is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Is the following argument valid: 'If it doesn't rain we'll have a picnic. We don't have a picnic. Therefore it rained.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The argument is valid because the only condition for not having a picnic was that it would rain, since there was no picnic, it must have rained. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In this argument where told the something must happen as a result of a certain condition. That thing is not happen, so the condition cannot have been satisfied. The condition was that it doesn't rain; since this condition cannot have been satisfied that it must have rained. The argument is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. We can symbolize the following argument: 'If it rains, the grass gets wet. If the grass gets wet, we'll be able to smell the wet grass. It rained yesterday. Therefore yesterday we were able to smell the wet grass.' Let p stand for 'It rains', q for 'the grass gets wet' and r for 'we can smell the wet grass'. Then the first sentence forms a compound statement which we symbolize as p -> q. Symbolize the remaining statements in the argument. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q, q r (p q) ^ (q r) I understand the equation up till this point [ (p -> q) ^ (q -> r) ^ p] -> r. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The argument gives three conditions, 'If it rains, the grass gets wet. If the grass gets wet, we'll be able to smell the wet grass. It rained yesterday.', which are symbolized p -> q, q -> r and p. It says that under these three conditions, the statement r, 'we can smell the wet grass', must be true. Therefore the argument can be symbolized by the complex statement [ (p -> q) ^ (q -> r) ^ p] -> r. STUDENT COMMENT: becuase the statment is valid r will be on the outside of the parenthesis INSTRUCTOR RESPONSE: It doesn't matter whether the statement is valid or not. The premises go into the parentheses or brackets, the conclusion follows the -> sign. The form of the argument is [premises] -> conclusion, where the premises inside the brackets are joined by conjunctions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. The preceding argument was symbolized as [ (p -> q) ^ (q -> r) ^ p] -> r. Determine whether this statement is true for p, q, r truth values F F T. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For these truth values p -> q is true since p is false (recall that the only way p -> q can be false is for p to be true and q to be false), q -> r is false since q is false, and p itself is false, therefore [ (p -> q) ^ (q -> r) ^ p] is false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For these truth values p -> q is true since p is false (recall that the only way p -> q can be false is for p to be true and q to be false), q -> r is false since q is false, and p itself is false, therefore [ (p -> q) ^ (q -> r) ^ p] is false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. ********************************************* Question: `q007. Symbolize the following argument: If it rains, it pours. It doesn't rain. Therefore it doesn't pour. Then set up a truth table to test the validity of the argument. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (p q) ^ (~p ~q) p q ~p ~q (~p ~q) T T F F F T F F T F F T T F T F F T T T confidence rating #$&*:nestly, I am just guessing by this point….sorry ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: "