Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

I think that the question is contained in the title.
Suppose we begin from something that is false for example $1=0$.
Is it possible using only $\Rightarrow$ (and of course $\lnot ,\wedge,\lor$) to prove any possible statement?
It has been ages since i studied logic at the university so i would prefer a simple explanation
(if possible)

share|improve this question

marked as duplicate by ShreevatsaR, M Turgeon, hardmath, Brian Rushton, Sami Ben Romdhane Dec 23 '13 at 14:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Related: Why is it possible to conclude everything from a false statement? (which is itself a duplicate of two other questions, but has a good answer there, so I'm linking to it). –  ShreevatsaR Dec 23 '13 at 13:49
1  
xkcd.com/704 :) –  Doorknob 冰 Dec 23 '13 at 14:55
    
Relevant: math.stackexchange.com/questions/148210/… –  MJD Dec 29 '13 at 1:58

4 Answers 4

up vote 10 down vote accepted

If we accept a false statement, can we prove everything?

If by "false" we mean contradictory, then the answer is yes, by the principle of explosion. For example, begin with the Peano axioms and adjoin the sentence $1=0$. Then since it is already a theorem of the Peano axioms that $1 \neq 0$, we have a contradiction. We say that the resulting system is "inconsistent" (meaning, it has a contradiction), and therefore satisfies the conditions under which the principle of explosion can be invoked.

Now on the other hand, if by "false" we mean unsound, then the answer is not necessarily. For example, begin once again with the Peano axioms, but this time adjoin an additional axiom that says "the Peano axioms are inconsistent." Paradoxically, this does not entail a contradiction. Therefore, we cannot prove everything. For example, pick any sentence $\varphi$ in the language of the Peano axioms; we cannot prove both $\varphi$ and $\neg \varphi$ in the system under discussion. Nonetheless, we accepted a statement (namely, that the Peano axioms are inconsistent) which is almost surely false.

share|improve this answer

If you are using Natural Deduction (rif.Ian Chiswell & Wilfrid Hodges, Mathematical Logic, 2007), - pag.25 - you have the ($\lnot$-intro) rule (or RAA : Reductio Ad Absurdum) :

suppose we have a derivation $D \Rightarrow \bot$ whose conclusion is $\bot$ , then there is a derivation $D' \Rightarrow \phi$ whose assumptions are those of $D$, except possiby $(\lnot \phi)$ (where - pag.24 - $\bot$ is a statement which is definitely false (absurdity)).

In a "traditional" approach, you will exploit the tautology : $A \rightarrow (\lnot A \rightarrow B)$ ; but you need a contradiction and not a false statement.

Only if you are working in a theory like first-order Peano Arithmetic, where you can prove that $1 \neq 0$ (i.e. $\vdash \lnot 1 = 0$ ), from your assumption $1=0$, with the above tautology, you can derive $B$ (whatever).

share|improve this answer

You can use a truth table to prove that $P\implies [\neg P\implies Q]$.

In words, if we know that $P$ is true, but we assume, for the sake of argument, that $P$ is false (i.e. $\neg P)$, then any proposition $Q$ can be derived.

We can also prove this theorem as follows:

  1. Suppose $P$

  2. Suppose $\neg P$

  3. Suppose $\neg Q$

  4. We obtain the contradiction $P\land \neg P$

  5. Therefore, we must have $\neg\neg Q$ or $Q$

  6. Therefore, $\neg P \implies Q$

  7. Therefore, $P\implies [\neg P\implies Q]$

share|improve this answer

Any theorem is based on axioms and established theorems. So when you add a deductive theorem to your system of theorems. You have to guarantee your system is self-consistent. Otherwise, the system is nonsense.

Of course you can add $1=0$ in your system of theorems, but I don't know what's your established system. In ring theory, $1=0$ implies a trivial ring and that's totally fine. But if $1=0$ is based on axioms of natural numbers, then you can derive anything since "$1=0$"$\to q$ is true whatever $q$ is because $1=0$ is false. But, certainly, what you derived from $1=0$ may contracts with established theorems or axioms in system of theorems.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.