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Determine whether the argument is valid or invalid.
Loretta's hobby is stamp collecting. If her husband likes to fish, then Loretta's hobby is not stamp collecting. If her husband does not like to fish, then Nathan likes to read. Therefore, Nathan likes to read.
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Let's introduce some pronumerals to make things simpler.
p = Loretta's hobby is stamp collecting.
q = Her husband likes to fish.
r = Nathan likes to read.
Then the statements read:
We know p is true. By taking the contrapositive of [1], we have:
...[3] (hmm...)Combining [2] and [3] using the law of transitivity, we have:
Therefore, the argument is valid.
Last edited by Identity (2007-10-30 18:38:29)
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Oh my god. I am learning more here than in school. Thanks so much i have a test in the mornin.
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Actually, I'm not 100% about my use of the contrapositive...
While it's true that if
I'm worried that this might not work if you have negate b:
soz, you could try looking up the properties of the contrapositive
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For the contrapositive of q then p :
not q then not p
Last edited by rune2402 (2007-10-30 18:48:51)
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I have to regroup tomorrow on this. I'll study transitivity and back here.
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This one is killing me because I really do not understand the law of transitivity or how it relates to the validity of an argument. I'll be studying this in depth. I'll be back.
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I'm worried that this might not work if you have negate b:
then
That is perfectly fine. It works.
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I am totally lost. it appears this question was the only one in my book section that i completely do not munderstand. I still o not know mif it is valid or not. I will have to regroup on this one.
You are a genius.
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Identity wrote:I'm worried that this might not work if you have negate b:
thenThat is perfectly fine. It works.
Yeah what she said. If you want to be really rigorous, you could include an intermediate step in there of
Why did the vector cross the road?
It wanted to be normal.
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