Sentential Logic: Is the argument valid?

  • Thread starter Thread starter Saladsamurai
  • Start date Start date
  • Tags Tags
    Argument Logic
Saladsamurai
Messages
3,009
Reaction score
7

Homework Statement



I am going over the first chapter of Velleman's How to Prove It. I have never studied logic before and while I understand most of what is happening, I am not sure that I am understanding how to transform sentences into symbols that can be tested. One of the exercises asks: Identify the premises and conclusion and determine if the following argument a valid one?

Jane and Pete won't both win the math prize. Pete will win either the math prize or the chemistry prize. Jane will win he math prize. Therefore Pete will win the chemistry prize.

Homework Equations



A valid argument is one in which the premises cannot all be true without the conclusion being trues as well

The Attempt at a Solution



I have identified the premises to be:
1.) "Jane and Pete won't both win the math prize",
2.) "Pete will win either the math prize or the chemistry prize",
3.) "Jane will win the math prize."

and the conclusion to be:
"Pete will win the chemistry prise."

Are my premises correct? Or should I have not listed (1) and (2) separately? That is, should I have combined them to say:
"Jane and Pete won't both win the math prize AND Pete will win either the math prize or the chemistry prize" ??

In order to test the validity, I need to construct a truth table.

I have translated my premises symbolically as follows:

Define the follwing symbols:

P: Pete will win the math prize
J: Jane will win the math prize
P': Pete will win the chemistry prize

##\lnot(P \land J)\land(P \lor P')##
##J##
##\therefore P'##

Now for the truth table:

<br /> \begin{array}{l c l} <br /> PJP&#039; &amp; [\lnot(P \land J)\land(P \lor P&#039;)] &amp; P&#039; \\<br /> ---&amp;------&amp;---\\<br /> TTT &amp; F &amp; T \\<br /> TTF &amp; F &amp; F \\<br /> TFF &amp; T &amp; F \\<br /> \dots &amp; &amp;<br /> \end{array}<br />I can already see that I am doing something wrong here. The last row has that my premise is true, but the conclusion is false, however, I know from the solution that the argument is valid. So where am I going wrong?
 
Physics news on Phys.org
it seems to me your premise(s) should be:

[~(P&J) & (PvP')] & J

which is all the premises you are given.
 
Hi Deveno :smile: Thank you for your response. Looking back, it seems obvious now that I should have included this. So, in general, if we have a list of premises leading up to some conclusion and we wish to test the validity, we should "string" all of the premises together into a single statement that can be tested. Does this sound like the general approach?

Thanks again.
 
Saladsamurai said:
Hi Deveno :smile: Thank you for your response. Looking back, it seems obvious now that I should have included this. So, in general, if we have a list of premises leading up to some conclusion and we wish to test the validity, we should "string" all of the premises together into a single statement that can be tested. Does this sound like the general approach?

Thanks again.

yes, but...

often, there are ways to "simplify" the premises, before conducting the validity test.

in terms of mathematical proofs, the premises are often "pre-conditions" like so:

suppose p is a prime integer > 2, then p+1 is even.

which has the premises:

A) p is in Z
B) p is in P (the prime numbers...insert favorite definiton here:_____)
C) p > 2

and conclusion:

D) p+1 is even

so the statement is A&B&C → D, but "A" is superfluous, then statement B&C → D is true whenever our original statement is, and vice-versa (because B is a stronger statement than A, that is: B → A).

the entire (mathematical) meaning of p→q is encapsulated in the phrase: "(p→q is valid (i.e.: true) if...) p cannot be true without q also being true".

in ordinary thinking, this means that "p" is a restriction to a perhaps more general concept ("q").
 
Deveno said:
yes, but...

often, there are ways to "simplify" the premises, before conducting the validity test.

in terms of mathematical proofs, the premises are often "pre-conditions" like so:

suppose p is a prime integer > 2, then p+1 is even.

which has the premises:

A) p is in Z
B) p is in P (the prime numbers...insert favorite definiton here:_____)
C) p > 2

and conclusion:

D) p+1 is even

so the statement is A&B&C → D, but "A" is superfluous, then statement B&C → D is true whenever our original statement is, and vice-versa (because B is a stronger statement than A, that is: B → A).

the entire (mathematical) meaning of p→q is encapsulated in the phrase: "(p→q is valid (i.e.: true) if...) p cannot be true without q also being true".

in ordinary thinking, this means that "p" is a restriction to a perhaps more general concept ("q").

An answer and then some! That's great Deveno. Thanks for your insight. :smile:

EDIT: Also, another way to fix my truth table would have been to simply add another column separately with J as a premise. Again, a result of the same oversight.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top