#$&* course Mth151 nov 4th9:43am 17. `query 17
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Given Solution: `a** This argument is an instance of the 'fallacy of the converse'. In commonsense terms we can say that there could be many reasons why he might be ecstatic--it doesn't necessarily follow that it's because he doesn't have to set up. A Venn diagram can be drawn with 'doesn't get up' inside 'ecstatic'. An x inside 'ecstatic' but outside 'doesn't get up' fulfills the premises but contradicts the conclusions. Also [ (p -> q) ^ q ] -> p if false for the p=F, q=F case. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.6.11 (formerly 3.6.12). This wasn't assigned but you should be able to analyze it. She uses ecommerce or uses credit. She doesn't use credit. Therefore she uses ecommerce. Is the argument and valid or invalid and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: disjunctive syllogism confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The argument can be symbolized as p V q ~q therefore p This type of argument is called a disjunctive syllogism. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.6.18 evaluate using the truth table: ~p -> q, p, therefore -q YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T T F F F .... F ..... T T F F T T .... F ..... T F T T F T .... T ..... F F F T T T .... T ..... T confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We need to evaluate {(p--> ~q) ^ ~p} --> ~q, which is a compound statement representing the argument. p q ~p ~q (p--> ~q) {(p--> ~q) ^ ~p} {(p--> ~q) ^ ~p} --> ~q then truth table is p q ~p ~q (p--> ~q) {(p--> ~q) ^ ~p} {(p--> ~q) ^ ~p} --> ~q T T F F F F T T F F T T F T F T T F T T F F F T T T T T Note that any time p is true (p->~q)^~p) is false so the final conditional (p->~q)^p) -> ~q is true, and if q is false then ~q is true so the final conditional is true. The F in the third row makes the argument invalid. To be valid an argument must be true in all possible instances. } Another version of this problem has ~p -> q and p as premises, and ~q as the consequent. The headings for this version of the problem are: p q ~p ~q ~p -> q (~p -> q) ^ p [ (~p -> q) ^ p ] -> ~q. Truth values: T T F F T T F T F F T T T T F T T F F F T F F T T T F T The argument is not true by the final truth value in the first line. To be true the statement [ (~p -> q) ^ p ] -> ~q must be true for any set of truth values. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q3.6.24 evaluate using the truth table: ( (p ^ r) -> (r U q)), and (q ^ p), therefore r U p YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ftt ... f , f , t , t fft ... f , f , t , t ttf ... f , t , t , t tff ... f , f , t , f ftf ... f, f , f , t fff ... f , f , f , f confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The argument ( (p ^ r) -> (r U q), and (q ^ p), therefore r U p is a conditional, with antecedent { ((r ^ p ) --> (rU q)) ^ (q^p) } and consequent (rUp). The argument can therefore be written as { ((r ^ p ) --> (rU q)) ^ (q^p) } --> (r U p) The headings can be set with the following headings: p, q, r, p^r, rUq, q^p, (p^r)->(rUq), {((r ^ p ) --> (rU q)) ^ (q^p)}, {((r ^ p ) --> (rU q)) ^ (q^p)} --> (rUp) This permits each column to be evaluated, once the columns for p, q and r are filled in by standard means, by looking at exactly two of the preceding columns. Here's the complete truth table. The table as shown below should, but may not, line up correctly. In any case you can go by the table above: pqr r^p q^p rUp rUq (r^p)->(rUq) [(r^p)->(rUq)]^(q^p) {[(r^p) -> (rUq)] ^ q^p} -> rUp ttt t t t t t t t tft t f t t t f t ftt f f t t t f t fft f f t t t f t ttf f t t t t t t tff f f t f t f t ftf f f f t t f t fff f f f f t f t All T's in the last column show that the argument is valid. COMMON BAD IDEA: p, q, r, (r ^ p), (rUq), (q^p), (rUp), {[(r^p)->(rUq)] ^ (q^p)}->(rUp) You're much better off to include columns for [(r^p)->(rUq)] and {[(r^p)->(rUq)] ^ (q^p)} before you get to {[(r^p)->(rUq)] ^ (q^p)}->(rUp). If you have to look at more than two previous columns to evaluate the one you're working on you are much more likely to make a mistake, and in any case it takes much longer to evaluate. ** (faulty) STUDENT SELF-CRITIQUE see above (no statement was included given in self-critique; statement 'I keep looking at this question and I still can't seem to understand. ' was included as student's solution but not in the self-critique.) INSTRUCTOR RESPONSE Your work has generally been very good, and you had good reason to be confused with the solution that was given on this question. You self-critique was however insufficient. Your statement should be included in your self-critique. Your reader shouldn't have to go back to look up information. Of course that's not really a problem because I remember your comment, but in general a comment should follow, not precede, the information to which it refers. You apparently understood the preceding problem, and if so you would be able to provide more specific information about what you do and do not understand about the given solution. In any case the solution to the problem has been revised. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q3.6.30: Christina sings or Ricky isn't an idol. If Ricky isn't an idol then Britney doesn't win. Britney wins. Therefore Christina doesn't sing. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ~r -> ~b is b -> r this is not valid due to the true table confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Solution using deductive reasoning: If r stands for RM is a teen idol c stands for CA sings b stands for BS wins then the statements are c U ~r ~r -> ~b b therefore ~c. The contrapositive of ~r -> ~b is b -> r. So we can rewrite our statement ~r -> ~b b as b -> r b from which we conclude that r is true. Combining this with our first statement c U ~r: c U ~r r from which we conclude by disjunctive syllogism that c is true. That is, Britney wins so Ricky is an idol. Christina sings or Ricky isn't an idol. So Christina sings. The argument concludes ~c, the Christina doesn't sing. So the argument is invalid Solution using truth tables: If we let p stand for Christina sings, r for Ricky is a teen idol and w for Britney wins AMA award then we have p V ~r ~r->~w w Therefore ~p The argument can be expressed as the statement [(pV~r)^(~r->~w)^w]-~p We can evaluate this statement using the headings: p, r, w, ~r, ~w, ~p, (pV ~r), (~r->~w), [(pV~r)^(~r->~w)^w] , [(pV~r)^(~r->~w)^w]-~p. We get (The lines below might not align correctly; the picture above is correctly aligned) p r w ~r ~w ~p (pV ~r) (~r->~w) [(pV~r)^(~r->~w)^w] [(pV~r)^(~r->~w)^w]-~p T T T F F F T T T F T T F F T F T T F T T F T T F F T F F T T F F T T F T T F T F T T F F T F T F T F T F F T T F F F T F F T T F T T T T T F F F T T T T T F T. The argument is not valid, being false in the case of the first row. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qPrevious version 3.6.30 determine validity: all men are mortal. Socrates is a man. Therefore Socrates is mortal YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (r -> p) ^ (p -> q) ] -> (r -> q) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This can be reasoned out by the transitive property of the conditional. If p stands for 'a man', q for 'mortal', r for 'Socrates' you have r -> p p -> q therefore r -> q which is valid by the transitive property of the conditional. A truth-table argument would evaluate [ (r -> p) ^ (p -> q) ] -> (r -> q). The final column would come out with all T's, proving the validity of the argument. ** STUDENT COMMENT I was saying p=men are mortal, q=Socrates is a man, and r= Socrates is mortal and so this is why I was having trouble. INSTRUCTOR RESPONSE p, q and r need to stand for simple statements. None of the statements you quote here is a simple statement. 'Socrates is a man' is not a simple statement. It means 'if it's Socrates, then it's a man'. Similarly 'Socrates is mortal' means 'if it's Socrates, then it's mortal'. The statement 'men are mortal' says 'if it's a man, then it's mortal'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!