course Mth 151 ?????a?????Q?l??·Fl??assignment #017
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12:05:26 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 --> It is invalid because of the fallacy of the converse. The argument assumes that {(p ->q) ^ q} -> p, but this isn't true. confidence assessment: 3
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12:05:31 ** 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 self critique assessment: 3
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12:07:51 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 --> This is valid by disjunctive syllogism: {(pUq) ^ ~q} -> p confidence assessment: 3
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12:07:58 ** 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 self critique assessment: 3
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12:11:21 Query 3.6.20 evaluate using the truth table: ~p -> q, p, therefore -q
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RESPONSE --> p q (p -> ~q) (p->~q)^~p [(p->~q)^~p]->~q T T F F T T F T F T F T T T F F F T T T This is not valid. confidence assessment: 3
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12:11:30 ** 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 self critique assessment: 3
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12:15:32 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 --> p q r (p^r)->(rUq)^ q^p (p^r)->(rUq)^ q^p -> r T T T T T T T F T T T F T F T T F F F T F T T F T F T F F T F F T F T F F F F T This is valid. confidence assessment: 3
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12:15:42 ** 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 self critique assessment: 3
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12:24:26 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 --> Christina sings = p Ricky isn't an idol= q Britney doesn't win= r p U q q -> ~r ~r Therefore ~p p q r (pUq) (q->r) ~r [(pUq)^(q->r)^~r T T T T T F F T T F T F T F T F T T T F F T F F T T T T F T T T T F F F T F T F T F F F T F T F F F F F F T T F [(pUq)^(q->r)^~r] -> ~p T T T F T T T T Because there is a False in the final column, this is invalid. confidence assessment: 3
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12:24:40 ** 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 self critique assessment: 3
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12:25:53 Previous version 3.6.30 determine validity: all men are mortal. Socrates is a man. Therefore Socrates is mortal
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RESPONSE --> All men are mortal = p Socrates is a man= q p q Therefore, p^q confidence assessment: 3
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12:26:31 ** 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 wasn't sure how to break the statements down and assign them letters, but now I think I understand. self critique assessment: 2
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