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Maybe a simple question, but I heard that an inconsistent theory can imply everything. For example: How to prove that $1=2$ from $0<0$.

user165633
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    $0<0$ is not inconsistent. It is inconsistent if you add it to a theory which proves $\lnot(0<0)$. But you didn't specify such a theory. – Asaf Karagila Jun 15 '15 at 12:21
  • @AsafKaragila Thanks, here I mean the usual theory of numbers. – user165633 Jun 15 '15 at 14:10
  • In the statement of your question you are mixing up the syntactic notion of consistency and the semantic notion of validity. Inconsistent theories imply everything by definition. Your example is an example of "ex falso quodlibet": the principle that anything follows from a single invalid statement. – Rob Arthan Jun 15 '15 at 21:51

2 Answers2

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The simple way:

Any statement of the type $$p\implies q,$$

where $p$ is equivalent to $False$, is by definition true. Also, the statement $0<0\wedge \neg(0<0)$ (equivalent to $False$), is true. Therefore, the statement $$(0<0 \wedge \neg(0<0)) \implies 1=2$$ is a true statement.

The more "intuitive" way:

It is simple to prove that $1\leq 2$, as this is true almost by definition.

You also know that if $ac \leq bc$ and $c>0$, then $a \leq c$.

Now, you know that $$0\leq 0\\ 2 \cdot 0 \leq 1\cdot 0$$ and because $0>0$, we can now assume that $2\leq 1$.

Therefore, we have $1\leq 2$ and $2\leq 1$, meaing that $1=2$.

5xum
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In standard mathematics we can prove $$ (b\cdot a = c \cdot a) \land (0<a) \implies b = c $$ In particular, by setting $a=0$, $b=1$, $c=2$ we get $$ 0=0 \land 0<0 \implies 1 =2 $$ The first premise $0=0$ is certainly true, so if we assume $0<0$ the conclusion $1=2$ must be true too.

hmakholm left over Monica
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