Okay.
Questions about an apparently simple argument.
- Thread starter ughaibu
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pka
Elite Member
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- Jan 29, 2005
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Here I find myself is disagreement with JeffM.
\(\left((q\to p)\wedge q\right)\to p\) is proposition not a argument for. Therefore it is true or false: Truth Table.
See the truth table tells us that the statement is true in all cases. Therefore it is known as a tautology.
Look at this argument form:
\(\begin{array}{*{20}{c}}
{1}&{q \to p} \\ {2}&q \\ \hline \therefore &p \end{array}\)
That is the valid argument form known as Modus Ponens.
\(\left((q\to p)\wedge q\right)\to p\) is proposition not a argument for. Therefore it is true or false: Truth Table.
See the truth table tells us that the statement is true in all cases. Therefore it is known as a tautology.
Look at this argument form:
\(\begin{array}{*{20}{c}}
{1}&{q \to p} \\ {2}&q \\ \hline \therefore &p \end{array}\)
That is the valid argument form known as Modus Ponens.
@ pka
I shall happily defer to you. This is one of your fields, not mine. In any case, I am done with this clown. I have tried very hard to understand what he was trying to get at and to make all this make some sort of intuitive sense to him. But what he wants to do is to shift his grounds from one silly argument to another. I also tutor at a language site and know what "cognate" means. "Mit der dummheit kämpfen götter selbst vergebens." And I live neither on Olympus nor in Asgard.