I just got introduced into arguments. How do I know if a certain argument is valid or invalid?

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- Nov 23rd 2011, 02:33 PMkmjtHow do I know if an argument is valid or invalid?
I just got introduced into arguments. How do I know if a certain argument is valid or invalid?

- Nov 23rd 2011, 02:39 PMPlatoRe: How do I know if an argument is valid or invalid?
- Nov 23rd 2011, 02:42 PMkmjtRe: How do I know if an argument is valid or invalid?
Oh i was talking in general. Heres an example, is it valid or invalid (i know answer just not sure why):

p --> (q --> r)

q --> (p --> r)

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Therefore (p or q) --> t - Nov 23rd 2011, 02:46 PMAnnatalaRe: How do I know if an argument is valid or invalid?
Generally speaking, I think you need to use a proof system (you should specify which one you're using) and one or more inference rules (modus ponens is most common).

When applied to sentential logic, a formula is valid iff it is satisfied by every interpretation. Or, put another way, it's a "tautology": a statement which must always be true regardless as to what we plug into the variables.

For example, the statement "not A or not B implies not (A and B)", under the usual (fixed) interpretation for not, and, or, and parentheses, will always be true regardless as to what the interpretation of A and B are.

Another example, in a more sequential fashion:

A

B

therefore...

A and B

Note that although the above argument is valid, it will still be false if one of its premises is false. Validity refers to the correctness of the logic, not the semantic interpretation. - Nov 23rd 2011, 02:50 PMAnnatalaRe: How do I know if an argument is valid or invalid?
When you say

**t**do you mean**T**(true)?*Anything*implies**true**, so that argument is valid no matter what the premises are. (The only time**A**does not imply**B**is when**A**is true, but**B**is false.) - Nov 24th 2011, 06:41 PMCaramelCardinalRe: How do I know if an argument is valid or invalid?
p --> (q --> r) == ~p v (~q v r)

q --> (p --> r) == ~q v (~p v r)

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Therefore (p or q) --> t == ~(p v q) v t

(p v t) == t (Universal bound law)