It would seem to me that if both and that p has nothing to do with q.
-Dan
Let us assume that we wish to prove something with a conjecture. We do not know whether it is true or false. But we assume that it is true and we arrive at our proof. After that we assume that it is false and we still arrive at the same conclusion. Thus we conclude that the thing we were proving must be true.
Here is the justification of the proof.
.
Thus, by the above paragraph:
(1)
(2).
Apply contrapositive to (1):
(3)
Apply syllogism (chain rule) between (2) and (3):
.
Assume is false thus,
is false thus contradiction.
Thus, is true.
Q.E.D.
But my question remains can you possibly think of something which you can prove like this? If you can this would be one elegant and amazing proof.
Not sure that is the case, for suppose (informal language being employedOriginally Posted by topsquark
so dont expect rigor) is "true" and , I think we don't have a problem
with this, we accept .
Now if we accept the law of the excluded middle, is "false", but a false proposition
implies every proposition so , but we still accept ?
If is "true" we just interchance and throughout.
RonL
I think that this reasoning is correct. However it could be surprising.
How we can conclude from p that q is true if we doesn't know either p is true or not???
It appears that we can. We have to realise that when we are prooving some theorem we use some axioms which are true for sure.
When p implies q and ~p implies q it means that:
1. The logic value of q is independent of the q logic value
2. The q is conclusion from axioms we used during proof process.
It shows us that there is no absolute truth in mathematical logic. Everyting depends on the axioms we assumed to be true (just as the value of the q).
Let denote axioms we assume to be true. Let be a conjecture and be our theorem. When we are prooving that p implies q (or ~p implies q) we are silently assuming that axioms are true. So we will have:
Let .
We will get:
Which means:
Which means that q is really conclusion from axioms.
Pretty long, huh. Ok, now back to work
I'm quite sure about that. The only way of showing that q is conclusion from p is to construct a proof based on some axioms. No matter if p is true or not if we show that p implies q (assuming p true) and than that ~p implies q (assuming the other possibility) there is no other way than to conclude that q is true (cause there is no more logical value p can have, "tertium non datur"). Proving the implications we use axioms so q is conclusion from axioms indeed.Originally Posted by CaptainBlack
If you have found some mistake please tell me about it.
To PerfectHacker: I haven't ever seen proof like this. I will try to think out if I have some time.
Sorry for 'prooving' in the previous post. My English isn't very good.
Goodnight
Sorry, but you missed my point. My comment was on the last sentence ofOriginally Posted by albi
you previous post:
I should have been more explicit. Are you sure that the axioms are true?It appears that we can. We have to realise that when we are prooving some theorem we use some axioms which are true for sure.
The point is truth is a slippery concept, most of the time we are satisfied
with asking if the axioms and the rules of proof are consistent.
But even then we usually only arrive at relative consistency. Which means
something like: System A is consistent if System B is consistent. System
B often being the system of Russel and Whitehead's Principia Mathematica.
But we have no proof of the consistency of PM.
RonL
CaptainBlack:
Yes, I do agree with you that 'truth' is slippery concept. My point is that (just as I have written) there is no absolute truth. If we imagine the class of all propositions (I don't know if I have used the propper word) we can see some relations between them, but none of them will be true unless we assume some of them are true (it is that relativeness you are talking about).
If I'm wrong, please correct me. I have never read Kurt Godel's works or even a good mathematical logic book.