B Help on this truth table- Is this statement valid

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The discussion revolves around constructing a truth table to assess the validity of a logical argument concerning Jane and Pete winning prizes. The original poster expresses uncertainty about their truth table, noting that not all premises are satisfied in their current version. Participants clarify that with three inputs (Jane, Pete, and Chemistry), there are eight possible cases to consider, and some can be eliminated based on the premises. They emphasize the importance of structuring the table with clear column headers for inputs, premises, and derived results, ultimately concluding that the argument is valid when analyzed as two interconnected syllogisms. The conversation highlights the complexity of truth tables and the need for careful consideration of premises and conclusions.
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This isn't a homework problem, but I was going to post it in there as my question is similar to one that would be asked there. However there doesn't seem to be a logic section for it so I posted it here.
The question is to make a logic table to determine if the following statement is valid:

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

I've attached an image of my truth table, but I don't think it's correct as in none of the cases are all three of my premises true.
 

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You haven't considered all eight cases. There are three inputs: J, P and C so there are eight possibilities. There are four you haven't considered, and they include ##J\wedge \neg P\wedge C##.
 
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Excellent thank you. I feel like this covers them all, but what if there's something like A,B,C,D,E? Is there a way we can determine how many unique cases there will be?
 

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There will always be ##2^n## rows in a full truth table when there are ##n## different inputs.
Often some can be quickly ruled out and on a pragmatic basis left out of the truth table, based on the axioms provided. For instance, in this case the axiom ##J## (Jane wins the maths prize) allows us to rule out all the cases where ##J## is false, thus narrowing our focus to a truth table with only four entries.
 
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andrewkirk said:
Often some can be quickly ruled out and on a pragmatic basis left out of the truth table
I would not leave them out, just in case things get confusing as they did here.
 
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Ok thank you guys. It seems the most difficult part is what to put in the column headers as the premises and conclusions.
 
You could also use 1st order here instead of sentence logic.
 
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opus said:
Ok thank you guys. It seems the most difficult part is what to put in the column headers as the premises and conclusions.
I think there are three types of columns: inputs, premises and derived results.

In this problem, there are three inputs: C, J, P.

There are three premises: the three statements in the OP, which you have labelled with orange dots in your picture.

Derived results will be expressions involving the inputs. There is at least one essential derived result column, which is the conclusion that is to be proved. In this case that is C, which may be confusing because it is also an input. We could double-label a column as both input and conclusion, but I think it makes it clearer to have separate columns for C as input and C as conclusion.

The other derived result columns, if any, are for parts of the conclusion. They can be helpful if the conclusion is a big, multi-part statement and it's easier to work out parts of it first and then combine them. We don't need those here because the conclusion statement is atomic.

Having set up the columns, my approach is to delete (eg rule a line through) each row in which any of the premises are false.
If there is one or more row remaining that means it is possible to satisfy the premises. If not then the theorem has been proved to be vacuously true.
Otherwise we proceed to delete any remaining rows in which the conclusion is false.
If any rows remain after that then the theorem is proven. If not it is disproven.
 
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Thank you very much guys. Looks like there's a little more to this than I thought so I'll need to do some more reading!
 
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opus said:
This isn't a homework problem, but I was going to post it in there as my question is similar to one that would be asked there. However there doesn't seem to be a logic section for it so I posted it here.
The question is to make a logic table to determine if the following statement is valid:

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

I've attached an image of my truth table, but I don't think it's correct as in none of the cases are all three of my premises true.
The argument is valid. The 3-premise argument can be viewed as two 2-premise syllogisms, the conclusion from the first of which is an enthymemic premise of the second.

Let's look at your 3-premise argument as two 2-premise arguments:

First, we restate the 3-premise argument:

Premise: Jane and Pete won't both win the math prize.
Premise: Pete will win either the math prize or the chemistry prize.
Premise: Jane will win the math prize.
Conclusion: Pete will win the chemistry prize.

The as-yet-unstated first 2-premise argument is:

Premise: Jane and Pete won't both win the math prize.
Premise: Jane will win the math prize.
Conclusion: Pete won't win the math prize.

The conclusion is not one of the original 3 premises, but it is entailed by the confluence of a particular 2 of them; the first and third of them.

Next we use the conclusion from the first argument as a premise in the second:

Premise: Pete will win either the math prize or the chemistry prize.
Premise: Pete won't win the math prize.
Conclusion: Pete will win the chemistry prize.

The second 2-premise argument, using the conclusion from the first 2-premise argument as a premise, yields the same conclusion as the 3-premise argument.
 
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