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.

I have some questions like if $P$ then $Q, P$ therefor $Q$ for example, how can you tell from writing your truth table if therefor $Q$ is valid or invalid? I mean I know its true because Modus Ponens tells me it is but that doesn't really help on more complex issues like;

p∨q
r
r → ¬q
−−−−−−
therefore p

I can make a table but what am I looking for in it to show me therefore p is valid or invalid.

Thanks

As per conversation with amwhy is this an accurate reflection of what you are trying to explain? I can see that the column with all true R is also true. Therefore its valid!enter image description here

share|improve this question
    
Thank you both muchly handy :) –  user133149 Apr 13 at 12:24
    
I meant to highlight the P not the R btw, It was just getting late. –  user133149 Apr 14 at 2:31

1 Answer 1

up vote 7 down vote accepted

You need to check the following:

The argument is valid if and only if whenever you have a row in which (all) entries under the following columns evaluate to true,

$p\lor q$

$r$

$r\rightarrow \lnot q$

Then we must also have $p$ true.


This is equivalent to checking whether the statement $$[(p \lor q) \land r\land (r\rightarrow \lnot q)]\rightarrow p$$ is a tautology (i.e., whether the statement evaluates to true for every possible truth-value assignment given to $p, q, r$. If it is a tautology, then the argument is valid:

enter image description here


Can you see why the two approaches listed above are equivalent?

share|improve this answer
    
Yer, I think so :) I started working on a table though to see if there was a column in which all entries evaluated to true. and I couldn't see one. I made a column where Q = T R = T and P = T then RvQ would equal true, R would equal True but R --> not Q equales F doesn't it. –  user133149 Apr 13 at 13:28
    
Unless I should be evaluating like ((r -> notQ)->p) –  user133149 Apr 13 at 14:08
    
You look at when each (all) of the premises are true (and only when they are all true), to confirm that in those cases, p is also true. You don't need to look at when $p, q, r$ are all true unless under that assignment (T, T, T), ALL the premises are true. And under that assignment, you've shown that one of the premises fails. You look at only those truth-value assignments to p, q, r that make all the premises true. In each of those cases, you'll find that $p$ is true. –  amWhy Apr 13 at 14:13
    
OK sorry about the miss-communication. So when we have a row when all of the premises are true, doesn't matter which row in the table? at that stage you look to see if p is also true? –  user133149 Apr 13 at 14:40

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.