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The classical propositional logic admits the concept of admissible rule, and would like some examples of propositional calculus with the 'admissible rule', on wikipedia I don't quite understand...

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    $\begingroup$ Please expand on your question. What exactly are you asking? $\endgroup$ Mar 10, 2015 at 17:10

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A rule $R= \langle S_0, \ldots, S_n, S \rangle$ is said to be a derivable in a proof system $\mathcal T$, if for each instance $S_0,\ldots, S_n,S$ there is a deduction of $S$ from the $S_i$ be means of the primitive rules of $\mathcal T$.

That is to say, in this deduction the $S_i$ are treated as additional axioms.

If we can consider the natural deduction proof system, we can derive from the primitive rules the derived rule :

$A \land B \to C \vdash A \to (B \to C)$.

A rule $R$ is said to be admissible for $\mathcal T$, if for all instances $S_0,\ldots, S_n,S$ it is the case that :

if for all $i \le n \ \ \vdash S_i$, then $\vdash S$.


See :

A derivable rule is admissible, but not necessarily the other way around.

Consider next the law of double negation: $¬¬A \to A$. If it is derivable [in intuitionsitic logic], the instance with $A ∨ ¬A$ in place of $A$, i.e., $¬¬(A ∨ ¬A) \to A ∨ ¬A$, is in particular derivable. [I.e.] if $¬¬ A \to A$ is derivable in intuitionistic logic, also $A ∨ ¬A$ is. Thus, that step constitutes an admissible rule.

It is a logical fallacy to conclude from such admissibility that the implication $(¬¬A \to A) \to (A∨¬A)$ is derivable.

We have that the admissible rules are those inference rules which can be consistently employed in derivations in a given system, i.e. they preserve the set of provable theorems.

We say that a system is structurally complete when the classes of admissible and derivable rules coincide; these systems are in some sense "self-containded" : their deductive apparatus generates all inference rules consistent with them.

In this sense, classical propositional calculus is structurally complete.

See :

See Intuitionistic Logic, Ch.4.2 Admissible rules of intuitionistic logic and arithmetic, for further examples :

  • Harrop's rule [1960] :

if $\vdash (¬A → (B ∨ C))$, then $\vdash (¬A → B) ∨ (¬A → C)$

is admissible but not derivable, because $(¬A → (B ∨ C)) → (¬A → B) ∨ (¬A → C)$ is not intuitionistically provable.

  • Mints' rule :

if $\vdash ((A → B) → A ∨ C)$, then $\vdash ((A → B) → A) ∨ ((A → B) → C)$.

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  • $\begingroup$ To take another example, given $\vdash x=0$ (for free variable $x$), we can substitute $1$ for $0$ to get $\vdash 1=0$, a contradiction, and so derive $\vdash x=1$. Thus $\vdash x=0\implies\vdash x=1$ is admissible. However, $\vdash (x=0\to x=1)$ is not derivable because if it was we could substitute $0$ for $x$ to get $\vdash 0=1$. $\endgroup$ Mar 11, 2015 at 9:06

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