Suppose, we have statement A which satisfies: if A then -C, where -C is the negation of C. In addition, we have: if A and B, then C. Then, by using the truth table, I found that the combination of:if (if A then -C) and (if A and B, then C), then the negation of (A and B) is always true, I mean it is tautology. Now my question, how do I used a false statement (A and B), which I already prove it is wrong, in the logical induction to disprove itself?
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1I don't quite understand what you're asking? The reason this is true is because if we assume $A\implies \lnot C$ and $(A\land B)\implies C$, then $A$ and $B$ cannot both hold; if they did, then by the first assumption $\lnot C$ would hold, and by the second, $C$ would hold, a contradiction. And you didn't use $A\land B$ to prove anything, so I don't see the problem. – florence Aug 15 '16 at 11:29
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But I used A∧B to get C and without this, I would not reached to the contradiction which proved that A∧B must be false. So how does one use a false statement (A∧B which I showed it is false) to prove that the same statement is false? Normally, one should use a valid statement to reach a conclusion. – Isaacadel Aug 15 '16 at 17:37
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You did not use $A \land B$ to get C. You used $(A \land B) \implies C$ to get $(A \land B) \implies (C \land \lnot C)$. to get $\lnot (A \land B)$. That is a valid statement. You never used or ever assumed $A \land B$. – fleablood Aug 16 '16 at 06:27
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Instead of truth tables reason. $(A \land B) \implies C$ doesn't assume $(A \land B)$. $(A \land B) \implies A$ doesn't assume $(A \land B)$. $A \implies \lnot C$. Therefore $(A \land B) \implies (C \land \not C)$ doesn't assume $(A \land B)$. Therefore $\lnot(A \land B)$. No-where in that argument as $A\land B$ assumed. – fleablood Aug 16 '16 at 06:40
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Note the following theorem of classical logic: $(A\to B)\wedge (A\to\neg B)\to \neg A$. It says that if from a proposition $A$ I can infer a contradiction (both $B$ and $\neg B$) then necessarily $A$ is not the case. Which is a standard method of proving negation of same claim in science.
Using this theorem in your case you can prove that from (1) $A\to \neg C$ (which imply $(A\wedge B)\to \neg C$) and (2) ($A\wedge B)\to C$ the desired $\neg (A\wedge B)$ follows. Which does only say that whenever (1) and (2) are true $A\wedge B$ is false.
Gur Ismael
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1@HenningMakholm: Yes, but I think that unless explicitly mentioned by the asker it is pointless to bring that up. After all, there are also many other non-classical logics besides intuitionistic logic. =) – user21820 Aug 15 '16 at 14:35
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"(if A and B, then C)" does not at all imply "A and B", and so you did not use "A and B" to prove anything. If A and B, then C and not C. But actually (as you have proven) not ( A and B ). If you're wondering about the logical validity of your argument, it is simply because the deductive rules are truth-preserving, meaning that if you start with only true sentences then you deduce only true sentences in their respective contexts. Thus if you can ever deduce both a sentence and its negation in the same context (in this case you deduced "C and not C" under the assumption "A and B"), then the only possible situation is that the context never occurs (in this case you therefore can conclude "not ( A and B )").
I think your problem is that you don't understand contexts. I'll show you the proof of your example in a clear and intuitive style.
$A \to \neg C$. [Premise 1; assumed true]
$A \land B \to C$ [Premise 2; assumed true]
If $A \land B$: [This specifies a context, not a statement!]
$C$. [by Premise 2 and the context]
$A$. [by the context]
$\neg C$. [by Premise 1 and the previous line]
$C \land \neg C$. [This is a true statement in this context.]
$\neg( A \land B )$. [The context where "$A \land B$" is true can never occur.]
Note that everything you deduce in any context is true in any subcontext (that is why we could 'drag' in the two premises into the context where "$A \land B$" is true, but not the other way around. In particular, all the things we deduced in the above proof in the context where "$A \land B$" is true may not be true in other contexts!
Also go through every step of the deduction and convince yourself that it is truth-preserving. For instance if we have deduced up to line 5, then "$A$" is true in the context where "$A \land B$" is true, and hence $\neg C$ must also be true in that context because of Premise 1. Therefore line 6 is valid.
Thus when we manage to prove "$C \land \neg C$" in the context where "$A \land B$" is true, we know by truth-preservation that "$C \land \neg C$" is a true sentence in the context where "$A \land B$" is true. Since we also know that "$C \land \neg C$" is always false in any context, we can conclude that "$A \land B$" can never be true. This also means that everything we proved under "If $A \land B$" is useless.
If you know programming this concept is trivial to understand, because when you have the code if(P){ ... } and P evaluates to false then the code inside the if-structure is never executed.
By the way, the style of proof shown above is a slight variant of Fitch-style natural deduction.
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So following the truth-preserving principle, I found myself starting with a statement that I proved to be wrong namely, "A and B" to imply C which when I used it and -C, I reached to "not (A and B). So how come I use a context which never occurs to deduce that this context never occurs? Is it possible to start with a context that never occurs to prove anything, even if this "anything" is the context that never occurs itself? – Isaacadel Aug 15 '16 at 17:55
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@Isaacadel: What is wrong with analyzing any context you like? We never assumed that the context was possible, but simply deduced what must be the case if the assumption held. You can't ignore the context of a statement. You didn't use "A and B" to deduce anything. Rather, under the assumption of "A and B" you can deduce "C and not C". Without that assumption you can't! What you deduce under an assumption may not be true outside it! So nothing is wrong. We see that the assumption cannot hold because we would be able to deduce a false sentence under it. – user21820 Aug 16 '16 at 04:57
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@Isaacadel: I've added further explanation to my answer. See if you get it now. – user21820 Aug 16 '16 at 06:09
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I can discuss and consider "If pigs poop gold than the planet mercury is a turnip" without assuming "pigs poop gold" is true. If pigs poop gold then my cousin would be rich. If pigs poop gold no-one would eat pork. If pigs poop gold I'd still have to go to work. These are all legitimate to say. As is if pigs poop gold my cousin would be rich. But my cousin is poor. Therefore pigs do not poop gold. All legitimate. We never said pigs do poop gold; just IF pigs did poop gold then... – fleablood Aug 16 '16 at 06:45
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@fleablood: Haha nice one! But eventually I think all explanations will fail unless one very carefully thinks about the meaning of "if P then Q" (as a conditional claim), because there's no non-circular explanation. – user21820 Aug 16 '16 at 06:55
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@Isaacadel: See fleablood's example in his above comment for a concrete example of the explanation in my answer. – user21820 Aug 16 '16 at 06:56
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"all explanations will fail unless one very carefully thinks about the meaning of "if P then Q" (as a conditional claim), because there's no non-circular explanation" Not sure I understand. Well maybe I do. Suppose we have "if P then Q" that doesn't say anything about P. But it does so "not(P and not Q)". If we also have "not Q" so we have "(P and not Q) or (not P and not Q)" but we have "not(P and not Q)" so we have "not P and not Q" and therefore "not P". Yeah, I can see a bit circularity there. But we never assumed anything about P until the very end where we proved "not P". – fleablood Aug 16 '16 at 07:14
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@fleablood: The circularity comes in the explanation of the meaning of "if". Notice that you used "if" in your explanation when you said "If we also have ...". My point was just that if one doesn't understand "if" then one can never understand it. (Pun completely intended.) But thankfully everyone does, though one has to think slowly what it says. – user21820 Aug 16 '16 at 07:19
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@fleablood: I did not follow what you said regarding proving not P. I don`t see tautology in not(P and not Q)? so we can not say "(P and not Q) or (not P and not Q)". – Isaacadel Aug 20 '16 at 15:48
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@Isaacadel: Don't worry about his later comment. Just make sure you understand his example as a concrete instance of the reasoning in my answer. – user21820 Aug 20 '16 at 16:10
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If we know "if P then Q" then we know "P and not Q" is impossible. So we know "not(P and not Q)". If we always know "P or not P". If we also know "not (Q)" then we know "(P and not(Q)) or (not P and not Q)". We already know "P and not Q" is impossible. So that only leaves "(not P and not Q)". Which means "not P". We have proven "not P" without ever assuming anything about "P". – fleablood Aug 20 '16 at 16:13
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Suppose we know "I am not made of spaghetti". But we don't know "it is raining". Well we do know that "(it is raining) or (it is not raining)" and as it doesn't matter whether it is raining or not I'm not made of spaghetti either way, we know "(it is raining AND I am not made of spaghetti") OR (it is not raining AND I am not made of spaghetti)". Now suppose some magical allknowing honest being came down and told you "If it is raining then fleablood is made of spaghetti" we'd know "it is raining AND I am not made of spaghetti" is impossible. – fleablood Aug 20 '16 at 16:22
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Replace "I am not made of spaghetti" with "I am not inside drinking hot cocoa" and the all-knowing host being saying "if it is raining then fleablood is inside drinking hot cocoa" and it will make perfect sense. – fleablood Aug 20 '16 at 16:25
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I am not sure if I have understood your question. However this is a possible explanation.
We have that $$\mbox{$(A\wedge B)\Rightarrow A\Rightarrow \neg C \quad\mbox{and}\quad (A\wedge B)\Rightarrow C$},$$ which implies that $(A\wedge B )\Rightarrow (\neg C \wedge C)=$False, and therefore $(A\wedge B)$ is False.
As a matter of fact, once we know that $(A\wedge B)$ is False, we can also say that $(A\wedge B)\Rightarrow (\mbox{whatever you want})$!! This not a paradox: if the hypothesis is always false you can not deduce anything from it, and so you can put whatever you want in the thesis.
Robert Z
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You said: "if the hypothesis is always false you can not deduce anything from it, and so you can put whatever you want in the thesis", do you mean: "you can" instead of "you can not". What is then the difference between (A and B)->(anything) and (hypothesis) can deduce (thesis)? – Isaacadel Aug 15 '16 at 18:02
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@Isaacadel If you have an implication $P\Rightarrow Q$ you deduce that $Q$ is true if you have that $P$ is true. But if you know that $P$ is always false then you can not say anything about $Q$ . So $Q$ can be whatever you want. – Robert Z Aug 15 '16 at 18:14