In propositional logic, if $\Gamma \vdash \bot$, does it mean I can't derive anything else from it?

Question

What I've been told is that an inconsistent set of propositions is irrelevant in propositional logic because if it derives a contradiction, then you can't derive anything else from it, but here's my doubt:

Take $\Gamma = \{p \land q, \neg(p \land q)\}.$ This clearly derives a contradiction. But by $(\land E)$, can't I derive both p and q from $p \land q$, such that I can say that $\Gamma \vdash p$ and $\Gamma \vdash q$?

From a contradiction, you can derive everything, not nothing. Who told you you couldn't derive anything? You just can't derive anything useful, because you can derive both $p$ and $\lnot p.$ — Thomas Andrews, Apr 18 '23 at 11:57
From $\bot$ we can derive a formula whatever, provided that Ex Falso rule : $\bot \vdash \varphi$ is part of the system. — Mauro ALLEGRANZA, Apr 18 '23 at 11:57
Essentially, we want "derivation" to mean truth, but if you can derive a contradiction, then everything is both true and false, which is not a useful logical system. — Thomas Andrews, Apr 18 '23 at 12:00

Tankut Beygu · Answer 1 · 2025-03-05T01:56:30.713

In most of the deduction systems, the principle of ex falso sequitur quodlibet is built into the definitions of the systems.

Let us see that

$$\Gamma\vdash\bot\implies\Gamma\vdash\phi$$

for any formula $\phi$ in a Hilbert-style axiomatic system where the principle is not immediately visible. We shall use Łukasiewicz's axioms

(Ax1)$\quad A\rightarrow(B\rightarrow A)$

(Ax2)$\quad(A\rightarrow(B\rightarrow C))\rightarrow((A\rightarrow B)\rightarrow(A\rightarrow C))$

(Ax3)$\quad(\neg A\rightarrow\neg B)\rightarrow(B\rightarrow A)$.

with the inference rule of modus ponens (MP). We shall employ essentially the metatheorems

$(A\rightarrow B)\rightarrow(\neg B\rightarrow \neg A)$, which is the converse of Ax3.
$\neg\neg A\rightarrow A$

The full proofs of these theorems would take us far afield; to have an idea, see, for example, my answer to a previous question).

Recall that the set of connectives $\{\rightarrow, \neg\}$ is functionally complete. Thus, we can express any formula with the familiar canonical connectives using the definitions

$A\wedge B\equiv\neg(A\rightarrow\neg B)$
$A\vee B\equiv\neg A\rightarrow B$
$A\leftrightarrow B\equiv\neg((A\rightarrow B)\rightarrow\neg(B\rightarrow A))$

We shall abbreviate a contradiction (e.g., $A\wedge\neg A$) with $\bot$ in the following argument; ipso facto, $\neg\bot$ is a theorem and can be written as a line by itself. So,

$\Gamma\vdash\bot\tag{Assumption}$
$\vdash\bot\rightarrow(\neg\phi\rightarrow\bot)\tag{Ax1}$
$\Gamma\vdash\neg\phi\rightarrow\bot\tag{MP 1, 2}$
$\vdash\neg\bot\tag{Theorem}$
$\vdash(\neg\phi\rightarrow\bot)\rightarrow(\neg\bot\rightarrow\neg\neg\phi)\tag{Metatheorem}$
$\Gamma\vdash\neg\bot\rightarrow\neg\neg\phi\tag{MP 3, 5}$
$\Gamma\vdash\neg\neg\phi\tag{MP 4, 6}$
$\vdash\neg\neg\phi\rightarrow\phi\tag{Metatheorem}$
$\Gamma\vdash\phi\tag{MP 7, 8}$

In propositional logic, if $\Gamma \vdash \bot$, does it mean I can't derive anything else from it?

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