# If we accept a false statement, can we prove anything? [duplicate]

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)

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.

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 equivalently $$Q$$

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

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

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).

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.