# Modus Ponens

• Sep 25th 2009, 12:40 AM
bobred
Modus Ponens
Hi

Here we go, we are told the first two propositions are true. We're asked is it valid and if so give a deduction by modus ponens and proof by contradiction.

(i)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave cannot solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is not quadratic.

(ii)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave can solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is quadratic.

This is probably really simple but I just keep going around in circles, can anyone explain?

Thanks
• Sep 25th 2009, 02:24 AM
Plato
Quote:

Originally Posted by bobred
We're asked is it valid and if so give a deduction by modus ponens and proof by contradiction.
(i)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave cannot solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is not quadratic.

(ii)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave can solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is quadratic.
This is not correct. You cannot make any conclusion here.

• Sep 25th 2009, 02:36 AM
bobred
Hi

Sorry for being vague, we are asked which of the two is a valid deduction.
For the valid one give a deduction involving modus ponens and proof by contradiction. For the invalid one explain why not.

Thanks
• Sep 27th 2009, 10:27 AM
bobred
Can anyone help?

Thanks
• Sep 27th 2009, 11:19 AM
Plato
Quote:

Originally Posted by bobred
Can anyone help?

The first is valid.
The second is invalid. The fallacy is call affirming the consequent.
• Sep 27th 2009, 11:55 AM
I agree with the above, but I'll add a bit more to illustrate Plato's point

Quote:

We know that:
If the equation is quadratic, then Dave can solve it;
Dave can solve the equation f(x)=0.
What if f(x) is, for example, a linear equation? If Dave can solve the equation f(x) = 0, where f(x) is a linear equation, then it is (obviously) not quadratic.

Arguement

X implies Y
Y
--
X <-- false. There could be a third cause, say, Z for Y.