They are not valid by their structure. No one has said that except you. I have given you examples of arguments that are not valid in that structure.
Questions about an apparently simple argument.
- Thread starter ughaibu
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My apologies for not getting it, can you give me an example of something on the lines of (Q∧R)→P that is not valid, please.
What I'm asking for is an example of invalid structure.
I'm asking for is an example of invalid structure
So, what is the invalid structure?
The invalid structure is
[MATH]q \land r \longrightarrow p.[/MATH]
And one way to demonstrate that it is an invalid structure is to give an example where true premises, in this case that 2 is a prime and that 7 is a prime, result in a false conclusion, namely that 7 + 2 = 9 = 3 * 3 is a prime.
So, q∧r⟶p. is invalid.
Yes
Yes. Here is one.
[MATH]((m \longrightarrow n) \land m) \longrightarrow n.[/MATH]
Here is what it means in English:
If it is true that n is true whenever m is true and also m is true, then n is true.
Notice that nothing is said about what happens if either of the conditions precedent is false.
[MATH]((m \longrightarrow n) \land m) \longrightarrow n.[/MATH]
Here is what it means in English:
If it is true that n is true whenever m is true and also m is true, then n is true.
Notice that nothing is said about what happens if either of the conditions precedent is false.
The structure??? I said one structure was not valid. I gave an entirely different structure that was valid.
Dr.Peterson
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I think you may be simply confusing "statement" with "argument". What you have here is merely a statement -- an assertion that q and r entails p, without proof. An argument is a series of statements (premises and a conclusion) that is intended to yield a true conclusion whenever its premises are true. You have to be able to demonstrate this (e.g. by a truth table) in order to call it valid.
A valid argument, for example, might be "q; r; q∧r⟶p; therefore p". Do you see the difference? If q and r are true, and the conditional statement is true, then p must be true.
Okay, please quote the structures side by side and point out the relevant difference.
[MATH](q \land r) \longrightarrow p[/MATH] is invalid.
[MATH]((q \longrightarrow p) \land q) \longrightarrow p[/MATH] is valid.
They are OBVIOUSLY not the same. For one thing, the first one has three different primitives indicated, p, q, and r, whereas the second has two primitives, p and q. For another thing, the first shows a single inference. The second shows two inferences.
[MATH]((q \longrightarrow p) \land q) \longrightarrow p[/MATH] is valid.
They are OBVIOUSLY not the same. For one thing, the first one has three different primitives indicated, p, q, and r, whereas the second has two primitives, p and q. For another thing, the first shows a single inference. The second shows two inferences.
It could be. It's nearly 7 in the morning and I'm quite drunk. I'll reply tomorrow.
Sure, but where are the cognates in my posts?