How do I prove p for p->q ?
Do I simply apply modus tollens then apply negation?
So
1) p->q
.....
2) ~q ...... Assumption
3) ~p ...... 1,2, modus tollens
4) p ..... 3, negation
Is that it?
BadgerJelly » September 25th, 2017, 2:01 am wrote:How do I prove p for p->q ?
BadgerJelly » October 1st, 2017, 7:05 am wrote:It's a conditional proof.
BadgerJelly » October 1st, 2017, 1:06 pm wrote:I messed up!
Should be ...
1- p -> q
2- ~q
----
3- p ........ assumption
----
4- q
5- ~q
Therefore ~p (reductio ad absurdum), but also the same thing in this instance as Modus Tollens (denying the consequent).
Users browsing this forum: No registered users and 5 guests