#$&* course MTH 151 10/20 around 11:45 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: It is possible that p is true and q is true. Another possibility is that p is true and q is false. A third possibility is that p is false and q is true. A fourth possibility is that p is false and q is false. These possibilities can be listed as TT, TF, FT and FF, where it is understood that the first truth value is for p and the second for q. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q002. For each of the for possibilities TT, TF, FT and FF, what is the truth value of the compound statement p ^ q ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p^q is a conjunction and must not contain any false statements to be true tt=true tf=false ft=false ff=false confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q003. Write the results of the preceding problem in the form of a truth table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q p^q t t t t f f f t f f f f confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The truth table must have headings for p, q and p ^ q. It must include a line for each of the possible combinations of truth values for p and q. The table is as follows: p q p ^ q T T T T F F F T F F F F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q004. For each of the possible combinations TT, TF, FT, FF, what is the truth value of the proposition p ^ ~q? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p for each statement is as stated tt= q would be negated and therefore be false and would make the conjunction false tf= ~q would be true, the conjunction would be true ft= p is false and ~q is false so the sonjunction is false ff= p is false even though ~q is true the conjunciton is still false confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q005. Give the results of the preceding question in the form of a truth table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q ~q p^~q t t f f t f t t f t f f f f t f confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The truth table will have to have headings for p, q, ~q and p ^ ~q. We therefore have the following: p q ~q p^~q T T F F T F T T F T F F F F T F &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Give the truth table for the proposition p U q, where U stands for disjunction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q pVq t t t t f t f t t f f f at least one statement must be true in order for the disjunction to be true confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. Reason out the truth values of the proposition ~(pU~q). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q ~q pV~q ~(pV~q) t t f t f t f t t f f t f f t f f t t f 1.tt so ~q=false. pV~q is true since p is true so its negagtion is false 2.tf ~q=true. pV~q is true since both are true. its negation is false 3.ft ~q=false. pV~q=false since both are false so its negation is true 4.ff ~q=true. pV~q=true since ~q is true so its negation is false confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the case TT p is true and q is true, so ~q is false. Thus p U ~q is true, since p is true. So ~(p U ~q) is false. In the case TF p is true and q is false, so ~q is true. Thus p U ~q is true, since p is true (as is q). So ~(p U ~q) is false. In the case FT p is false and q is true, so ~q is false. Thus p U ~q is false, since neither p nor ~q is true. So ~ (p U ~q) is true. In the case FF p is false and q is false, so ~q is true. Thus p U ~q is true, since ~q is true. So ~(p U ~q) is false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q008. Construct a truth table for the proposition of the preceding question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q ~q pV~q ~(pV~q) t t f t f t f t t f f t f f t f f t t f confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q009. Construct a truth table for the statement (p ^ ~q). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q ~q (p^~q) t t f f t f t t f t f f f f t f confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. Construct a truth table for the statement q U (p ^ ~q). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q ~q (p^~q) q V (p^~q) t t f f t t f t t t f t f f t f f t f f confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!