#$&* course MTH 151 014. Truth Tables
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Given Solution: It is possible that p is true and q is true. Another possibility is that p is true and q is false. A third possibility is that p is false and q is true. A fourth possibility is that p is false and q is false. These possibilities can be listed as TT, TF, FT and FF, where it is understood that the first truth value is for p and the second for q. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. For each of the for possibilities TT, TF, FT and FF, what is the truth value of the compound statement p ^ q ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TT- TRUE TF-FALSE FT- FALSE FF-FALSE confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Write the results of the preceding problem in the form of a truth table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTT TFF FTF FFF confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The truth table must have headings for p, q and p ^ q. It must include a line for each of the possible combinations of truth values for p and q. The table is as follows: p q p ^ q T T T T F F F T F F F F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. For each of the possible combinations TT, TF, FT, FF, what is the truth value of the proposition p ^ ~q? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TT-FALSE TF-TRUE FT-FALSE FF-FALSE confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Give the results of the preceding question in the form of a truth table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTFF TFTT FTFF FFTF confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The truth table will have to have headings for p, q, ~q and p ^ ~q. We therefore have the following: p q ~q p^~q T T F F T F T T F T F F F F T F &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Give the truth table for the proposition p U q, where U stands for disjunction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTT TFT FTT FFF confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Reason out the truth values of the proposition ~(pU~q). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TT- FALSE TF-FALSE FT-TRUE FF-FALSE confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the case TT p is true and q is true, so ~q is false. Thus p U ~q is true, since p is true. So ~(p U ~q) is false. In the case TF p is true and q is false, so ~q is true. Thus p U ~q is true, since p is true (as is q). So ~(p U ~q) is false. In the case FT p is false and q is true, so ~q is false. Thus p U ~q is false, since neither p nor ~q is true. So ~(p U ~q) is true. In the case FF p is false and q is false, so ~q is true. Thus p U ~q is true, since ~q is true. So ~(p U ~q) is false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. Construct a truth table for the proposition of the preceding question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTFTF TFTTF FTFFT FFTTF confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Construct a truth table for the proposition of the preceding question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTFTF TFTTF FTFFT FFTTF confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We need headings for p, q, ~q, p U ~q and ~(p U ~q). Our truth table therefore read as follows: p q ~q pU~q ~(pU~q) T T F T F T F T T F F T F F T F F T T F &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mth 151 Question: `q3.3.5 rewrite using if then ' all marines love boot camp '.
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Given Solution: `a** The statement is equivalent to 'If it's a Marine, it loves boot camp' or equivalent. The statement is not equivalent to 'if it is boot camp, then all Marines love it', which is the converse of the original statement. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q3.3.18 ~p false q false p -> q true YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: true confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Since ~p is false then p is true. Since q is false it follows that p -> q is of the form T -> F, which is false. The conditional is false when, and only when, the antecedent is true and the consequent false. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.3.36 write in symbols 'If we don't bike, then it does not rain.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p-r confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If p stands for 'don't bike' and r for 'it rains' then the statement would be p -> ~r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.3.48 q true, p and r false, evaluate (-r U p) -> p YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: false confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The antecedent (~r U p ) would be true, since ~r true and p false. The consequent p would be false. Since the antecedent is true and the consequent false, the conditional is false. ** MORE DETAILED SOLUTION r is said to be false, so ~r is true p is said to be false Therefore the disjunction (~r U p) would be a disjunction of a true and a false statement. A disjunction is true if at least one of the statements is true, so (~r U p) is true. The conditional (~r U p) -> p therefore consists of an antecedent which is true, and a consequent which is false. By the rules for a conditional, the statement is therefore false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.3.60 truth table for (p ^ q) -> (p U q) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tttt Tfftt Ftftt fffft confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The headings would be p, q ,(p^q), (pUq), (p^q)->(pUq) Row 1 would read T T T T T Row 2 would read T F F T T Row 3 would read F T F T T Row 4 would read F F F F T The common sense of this is that whenever both p and q are true, then the statement 'p or q' must be true. That's what means to say (p ^ q) -> (p U q). The fact that this statement is true is indicated by the last column of the truth table, which has True in every possible case. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.3.74 (formerly 3.3.72). This wasn't assigned but it is similar to assigned questions and should be answered: What is the negation of the statement 'if loving you is wrong then I don't want to be right' ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: LOVING YOU IS WRONG AND I WANT TO BE RIGHT confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The negation has to have the exact opposite truth values of the original statement. It is difficult and confusing to try to negate a conditional. It is much easier to translate the conditional to a disjunction then negate the disjunction. It is easy to negate the disjunction using deMorgan's Laws. Since p -> q is identical to ~p U q, the negation of p -> q is ~ ( ~p U q), which by de Morgan's Law is ~ ~p ^ ~q, or just p ^ ~q. So the negation would ge 'loving you is wrong AND I want to be right. COMMON ERROR AND NOTE: If loving you is wrong, then I want to be right. INSTRUCTOR COMMENT: The negation of a conditional can't be a conditional (a conditional is false in only one case so its negation would have to be false in three cases). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!