How do we test the validity of the following
(~p^ q) , (p--> r) , (~r-->s) , (s-->t) :- t
From 3 and 4 using law of syllogism , we can get~r-->t is true , but how do we test the validity of t only??
I don't think we can say anything about t
we know ~p but we have no information about ~p -> ??
Since ~p^q is true can we take ~p is true and because of that ~p-->~r is true(inverse of 2)
And then using the law of syllogism repetedly, we get ~p--> t is true.
And finally using law of detachment can we get t is true(~p-->~t is true and ~p is true)?
~p--> t is true*
I don't like it.
p -> r
does not mean that ~p -> ~r
The converse is not always true.
All we know is that ~r -> ~p
From #1, we know that ~p is true, but we know nothing about what happens for ~p
To test the validity of the given statements (~p^ q), (p--> r), (~r-->s), (s-->t) :- t, we can use the method of proof by contradiction. Here's how we can proceed:
1. Start by assuming the opposite of the conclusion, which is ~t.
2. From the last statement (s-->t) and our assumption (~t), we can conclude ~s using the contrapositive form of implication.
3. Now let's examine the third statement (~r-->s). Since we derived ~s in the previous step, we can conclude ~r using the contrapositive form of implication again.
4. Moving on to the second statement (p--> r), we can now conclude ~p using the contrapositive form once more.
5. Finally, let's consider the first statement (~p^ q). We derived ~p in the previous step, so we can conclude q from this statement.
Now, we have reached a contradiction. On the one hand, we concluded q from the first statement, and on the other hand, we concluded ~q from the assumption ~t. Since we arrived at a contradiction, our assumption ~t must be false, meaning t is true.
Therefore, we have shown that if ~(~p^q), (p-->r), (~r-->s), (s-->t) hold, then t must also hold. Thus, the given statements are valid.