Question:
I am confused about the use of temporary hypotheses. It seems dangerous to assume a statement is true, and then use that assumption to prove another statement. You have said on multiple occasions that from a wrong all else follows (1 + 1 = 1, therefore I am the Pope). If you make an error in the assumption, couldn't you prove an invalid statement true? For example, if you say that "Prove that for numbers in the domain of integers if a number is even it is odd"...(Vx)[A(x)->B(x)]. This statement is obviously false, but if you assume that all numbers are even (A(x) is true), the modus ponens tells us that all numbers are false. How can this be? In this case, we can make a false temporary hypothesis and prove a false statement true. I am certain there is a point that I am missng, which makes this problem go away, but I can't figure it out. Please give me some guidance that will allow me to use a temporary hypothsis without a high degree of skepticism.
Response:
You're right that it's dangerous to assume a random hypothesis - but the temporary hypothesis is far from random!
The point is that from
P
We add in
Q
and then (if we're lucky) we show that R follows. Hence
P and Q -> R
and using the deduction method backwards, we have that
P -> (Q -> R)
Now, if we'd assumed something absurd, something false, say, for Q, then all we're saying is that if P is true, then a tautology follows. The only case we're worried about is T -> F, and since Q -> R is always true (a tautology), we're not worried about this case!
In the end, of course, Q -> R should help us to move toward our goal of showing whatever it was to be shown, starting from P alone. (That's why I say that the temporary hypothesis was far from random.)
Does that help?
Response:
I am sure that I am missing something, because your explanation seems to prove my point. If I am asked to prove a false statement true, can't I just choose an incorrect temporary hypothesis that will make the statement true. In the majority of the problems we have done, we do not know what Q(x) is, so we choose the temporary hypothesis based on what will allow us to prove the statement true. I agree that if I am told a statement is true, a temporary hypothesis would not be random because only one assumption will prove the statement true. However, if I am told that a statement may, or may not be true, couldn't I choose a temporary hypothesis to prove the statement true, even if the statement were false? I still do not understand why, when the validity of the wff is unknown, the temporary hypothesis method is feasible.
Response:
I guess what you're missing is that it won't do you any good to add a tautology! You don't get to assume Q - you only get to assume that Q->R, and that's obvious since Q is false. Q -> R won't get you anywhere in a proof, since you'll never be able to assume Q!