Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is whether or not the following is a validly structured argument:

(P→T)→Q

Q → ¬Q

∴ P

I'm getting hung up on the Q→¬Q part by itself as a premise, it doesn't seem like that is permissible and I am unable to find an example of how to 'get rid of' the ¬ to prove Q. For example, modus tollens seems like it would be applied here, but I don't see how the truth of Q can imply that ¬Q is true. I'm not looking for an answer to the proof, just an explanation of how/why the premise is valid as stated.

share|improve this question
    
Unless I'm misunderstanding your notation, $Q\to\neg Q$ is a premise of the argument. The validity of the argument itself does not depend on whether or how you can establish the premises. If it happens that you can't prove the premises, then the argument is not of much use, but that doesn't mean that it's not, in itself, valid. –  Henning Makholm Apr 23 '12 at 23:09
    
@Henning: From $Q$ and $Q\to\neg Q$ you get an inconsistancy and thus, among everything else, $P$. –  Desiato Apr 23 '12 at 23:11
    
@Desiato: Right. Oops. –  Henning Makholm Apr 23 '12 at 23:12
    
Thanks for the responses. I can't say that I understand the why completely, but I'm assuming my understanding is way off from the start. –  logicalfallacy_Iam Apr 23 '12 at 23:24
    
@Desiato: but he doesn't have $Q$. From $Q \implies \lnot Q$ he can derive $\lnot Q$ –  Ross Millikan Apr 24 '12 at 0:36

2 Answers 2

up vote 1 down vote accepted

The premise $Q\to \neg Q$ is true not only when both $Q$ and $\neg Q$ are true, but also if $Q$ is false. In fact the second premise can be written as $$ Q\to \neg Q \equiv \neg Q \vee \neg Q \equiv \neg Q. $$ Therefore the second premise says that $\neg Q$ is true.

That doesn't tell how to get the conclusion, but it is a way to interpret the second premise. It turns out that the first premise is equivalent to $Q$ since $(P\to T)$ is always true, so the premises are contradictory. As mentioned in one of the comments, if there's a contradiction then all statements are both true and false, $P$ being one of those. Since any proposition is true based on these premises, one could just as easily use $\neg P$ as a valid conclusion.

share|improve this answer
    
Thank you, I can see why that makes sense. –  logicalfallacy_Iam Apr 24 '12 at 8:13

First of all, we can't ask the question of 'whether Premise is valid or valid'. Validity can only be ask of arguments.

You meant to question the intuitive sense behind the premise/proposition that 'If Q, then ~Q'.

At first glance, 'If Q, then ~Q' sounds crazy and makes no intuitive sense and downright contradictory.

If Q stands for it is raining, then 'If Q, then ~Q' stands for 'If it is raining, then it is NOT raining '. This sounds like a contradiction but it is not.

I guess your main concern is that 'If Q, then ~Q' sounds like a contradiction and so it is a confirmed false proposition and have no place in an argument.

Well, 'If Q, then ~Q' is not a contradiction.

Firstly, a proposition (compound statements) is make up of statements and a operator (such as -> or &, which signify the relationship between the variables)

A proposition is a contradiction if and only if no matter the truth value of the statements, the operator is false, meaning to saying the relationship between the statements is false. (i.e statements may be true or false but still, the relationship between the statements is still false.)

In the proposition ' If Q, then ~Q', Q and ~Q are your statements, If...then... is the relationship between these 2 statements.

This relationship is true, when statement Q is true and when ~Q is true. Meaning: If it is raining, then it is not raining.

So clearly, this proposition is not a contradiction since there is an instance whereby the relationship between the statements is true.

This relationship is true as well, when Q is false and when ~Q is false (~~Q is Q). Meaning: If it is not raining, then it is raining.

This relationship is true too, when Q is false and when ~Q is true. Meaning: If it is not raining, then it is not raining.

But, this relationship is false, when Q is true and when ~Q is false (~~Q). Meaning: If it is raining, then it is raining.

Thus, when i say 'If Q, then ~Q', i am not making a contradictory proposition. This proposition is true in 3 instances!

If you want to understand it intuitively, you can say:

" Well, it's all got to do with the if. If. if. if. If Obama wears glasses, he will look silly. In reality Obama is not wearing glasses, but IF he does, then he will look silly. You can't disprove my proposition since i am saying 'if'. Also, i can make a proposition 'if it is raining, then it is not raining.' I am saying 'IF'! If it so happens that Obama wears glasses and he does not look silly, then my proposition is wrong, but until then, i can make my proposition that If Obama wears glasses, he will look silly."

So, the conditional is really a special operator. Anyone can make any propositions using the conditional and get away with it until it is proven false in reality. Many philosophical debates involves trying to prove that the opponent's conditional premise is false in reality.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.