course Mth 151 017. `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):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qQuery 3.6.11 (formerly 3.6.12). This wasn't assigned but you should be able to analyze it. {}{}She uses ecommerce or uses credit. She doesn't use credit. Therefore she uses ecommerce. {}{}Is the argument and valid or invalid and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This type of argument is called a disjunctive syllogism confidence rating #$&* 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):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qQuery 3.6.18 evaluate using the truth table: ~p -> q, p, therefore -q YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q ~p ~q (p--> ~q) {(p--> ~q) ^ ~p} {(p--> ~q) ^ ~p} --> ~q T T F F F F T T F F T T F T F T T F T T F F F T T T T T confidence rating #$&* 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):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q3.6.24 evaluate using the truth table: ( (p ^ r) -> (r U q), and q ^ p), therefore r U p YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: this statement is true because the ts in the last columns match confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q3.6.30: Christina sings or Ricky isn't an idol. If Ricky isn't an idol then Britney doesn't win. Britney wins. Therefore Christina doesn't sing. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The argument is not valid, being false in the case of the first row confidence rating #$&* 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 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qPrevious version 3.6.30 determine validity: all men are mortal. Socrates is a man. Therefore Socrates is mortal YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the final column have all ts so the argument is valid confidence rating #$&* 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):ok ------------------------------------------------ Self-critique rating #$&*3