course Mth 151 ???O?????????assignment #017
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22:07:47 Query 3.6.6 if he didn't have to set up he would be ecstatic. He's ecstatic. Therefore he doesn't have to set up. Is the argument and valid or invalid and why?
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RESPONSE --> Invalid - fallacy of the converse ~p --> q q ~p
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22:08:09 ** 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 'no setup' inside 'ecstatic'. An x inside 'ecstatic' but outside 'no setup' fulfills the premises but contradicts the conclusions. Also [ (p -> q) ^ q ] -> p if false for the p=F, q=F case. **
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RESPONSE --> ok
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22:09:12 Query 3.6.12 she ecomms it or uses credit. She doesn't use credit. Therefore she orders it ecommsb Is the argument and valid or invalid and why?
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RESPONSE --> Valid - disjunctive syllogism p v q ~q p
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22:09:22 ** The argument can be symbolized as p V q ~q therefore p This type of argument is called a disjunctive syllogism. **
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RESPONSE --> ok
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22:13:06 Query 3.6.20 evaluate using the truth table: ~p -> q, p, therefore -q
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RESPONSE --> The ultimate answer is TTFT. I split my sections of the truth table as follows: p q (p --> ~q) ^ ~p --> ~q
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22:13:54 ** 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. **
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RESPONSE --> ok
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22:19:42 3.6.24 evaluate using the truth table: (p ^ r) -> (r U q), and q ^ p), therefore r U p
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RESPONSE --> Valid The final answer was TTTTTTTT. My columns were: p q r (r ^ p) --> (r v q) ^ (q ^ p) --> r v p
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22:20:19 ** The headings can be set up as follows: p q r p^r rUq (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. 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 ttf f t t t t t t tft t f t t t f t tff f f t f t f t ftt f f t t t f t ftf f f f t t f t fft f f t 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. **
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RESPONSE --> ok
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22:25:00 3.6.30: 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.
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RESPONSE --> Invalid. The final column is FTTTTTTT. My columns were: p q r (p v ~q) ^ (~q --> ~r) ^ r --> ~p
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22:25:37 ** 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 have b -> r b therefore r. We now have c U ~r r therefore c by disjunctive syllogism. That is, Britney wins so Rich 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 Martin is a teen idol and w for Britney Spears wins AMA award then we have p V ~r ~r->~w w Therefore ~p The argument is 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 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. **
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RESPONSE --> ok
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22:27:38 Previous version 3.6.30 determine validity: all men are mortal. Socrates is a man. Therefore Socrates is mortal
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RESPONSE --> I used an Euler diagram on this problem. My big circle was mortal men and the inner circle was Socrates. Since Socrates was in the mortal men circle, his is mortal.
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22:28:17 ** 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. **
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RESPONSE --> I guess I should have done a truth table, but I still got the correct answer.
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