course mth 151
I have also emailed this because it has a diagram in it.
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
017. `query 17
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `qQuery 3.6.5
if he didn't have to get up at 4 am he would be ecstatic.
He's ecstatic.
Therefore he doesn't have to get up at 4 am.
Is the argument and valid or invalid and why?
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Your solution:
It is valid because he is ecstatic when he doesn? get up a 4 am. He became ecstatic telling us he didn? have to get up.
You should include a description of your picture. What do your circles represent, which go inside of which, which overlap which, which do not overlap which, where do you put the X.
Confidence Assessment:
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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 'no getup' inside 'ecstatic'. An x inside 'ecstatic' but outside 'no getup' 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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Self-critique (if necessary):
In the book he is not just setting up as you described in the problem but that he is ecstatic if he doesn? get up at 4 am does this still make a fallacy?
There's a typo in my solution, s instead of g so what should read 'getup' was accidentally typed as 'setup'; the phrase should read 'no getup' inside 'ecstatic'.
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?
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Your solution:
valid because if she doesn? us credit the premise is either credit or commerce so she must of used e-commerce.
Confidence Assessment:
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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. **
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Self-critique (if necessary):
It didn? use the symbols
Self-critique Rating:
Question: **** `qQuery 3.6.18 evaluate using the truth table: ~p -> q, p, therefore -q
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Your solution:
p q ~p ~q ~p->q
T T F F T
T F F T T
F T T F T
F F T T F
invalid
Confidence Assessment:
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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. **
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Self-critique (if necessary):
I didn? evaluate it enough
you had a good start, but didn't put the argument together to evaluate it
Self-critique Rating:
Question: **** `q3.6.24 evaluate using the truth table: ((p ^ r) -> (r U q), and q ^ p), therefore r U p
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Your solution:
p q r p^q rVq q^p rVp
T T T T T T T
T T F T T T T
T F T F T F T
T F F F F F F
F T T F T F T
F T F F T F T
F F T F T F T
F F F F F F F
Confidence Assessment:
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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. **
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Self-critique (if necessary):
I am lost
you evaluated all the pieces of the argument, but didn't evaluate the argument itself. A couple more columns in your truth table and you would have had it.
The argument is
((p ^ r) -> (r U q), and q ^ p), therefore r U p. In symbols this is
( ( (p ^ r) -> (r U q)) ^ (q ^ p) ) -> (r U p).
Your truth table evaluated p q r p^q rVq q^p rVp
You would still need to a column for ( (p ^ r) -> (r U q)), then you would need a column for ( ( (p ^ r) -> (r U q)) ^ (q ^ p) ), and you would need a final column for ( ( (p ^ r) -> (r U q)) ^ (q ^ p) ) -> (r U p).
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.
Ashley Simpson sings or Ashton Kutcher is not a teen idol.
If Ashton Kutcher is not a teen idol, then Fantasia does not win a Grammy.
Fantasia wins a Grammy.
Therefore, Ashley Simpson does not sing.
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Your solution:
p= A.S. sing
q=
This is invalid because the statement is A.S. sings or A.K is not a teen idol.
If he is not a teen idol, then Fantasia doesn? win a Grammy.
But she does win a Grammy so that makes A.S sing.
Confidence Assessment:
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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, 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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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
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Your solution:
#36
This is valid but I can? explain why in symbols.
Confidence Assessment:
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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. **
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.
p, q and r need to stand for simple statements.
'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'.