Hypothesis-Discharging Rules in Atomic Bases

Abstract

This paper investigates the idea, familiar inter alia from Prawitz, that an inference is to be deemed valid, relative to a basis of inference rules for atomic sentences, just in case every extension of that basis supporting the premisses of the inference also supports its conclusion. Specifically, we try to carry out this idea in a setting where atomic bases are allowed to contain rules that license the discharging of hypotheses. The results are mixed. While the ensuing concept of validity appears to be an extensionally adequate one, from a conceptual point of view the theory is deemed unsatisfactory in that certain inferences come out valid, as it were, for the wrong reason. So, for instance, an atomic rule contained in a basis will qualify as valid relative to that basis, but not simply in virtue of the fact that it occurs there; its validity also depends on the assumption that all bases conform to a particular format, thus lending a peculiarly holistic character to the theory.

Appendix

The purpose of this Appendix is to substantiate the claim, made in Sect. 14.3, that, whether or not there exists such a thing as the smallest relation satisfying Clauses (20) and (21 $'$ ), any such smallest relation fails to satisfy (24).

Consider the following conditions (corresponding to (20), (21$'$) and (24), respectively) on an arbitrary three-place relation $\mathfrak X$ between sets of atomic sentences, type-2 bases, and individual atomic sentences.

• (R) If $\mathbf P \mathbin {\Rightarrow } q \in \mathcal B$, and $R \mathrel {{\mathfrak X}_{\mathcal B}} p$ for every $R \mathbin {\Rightarrow } p$ in $\mathbf P$, then $\mathrel {{\mathfrak X}_{\mathcal B}} q$.

• (E) If $\mathrel {{\mathfrak X}_{\mathcal C}} q$ for every $\mathcal C \supseteq \mathcal B$ such that $\mathrel {{\mathfrak X}_{\mathcal C}} p$ for every $p$ in $P$, then $P \mathrel {{\mathfrak X}_{\mathcal B}} q$.

• (M) If $P \mathrel {{\mathfrak X}_{\mathcal B}} q$ and $\mathcal C \supseteq \mathcal B$ then $P \mathrel {{\mathfrak X}_{\mathcal C}} q$.

We begin by stating that $\vdash$ is the smallest relation $\mathfrak X$ satisfying (R), (E), and (M) (where $\vdash$ is as defined in Sect. 14.4). That $\vdash$ does indeed satisfy the three conditions has been established in Sect. 14.4. And that $\vdash$ is included in any $\mathfrak X$ satisfying the conditions follows by induction from the fact that any such $\mathfrak X$ will also satisfy clauses (25) and (26), with $\mathfrak X$ in place of $\vdash$. The details are left to the reader.

Next, we show that there exists a relation $\mathfrak S$ which satisfies (R) and (E) yet does not include $\vdash$.

Such a relation $\mathfrak S$ can be constructed as follows. Pick two distinct atoms $a$ and $b$, and define, for any $\mathcal B$:

$$\begin{aligned} \begin{array}{l@{\quad }ccl} &{} {\mathfrak {S}}_{\mathcal {B}}\, b &{} {\Leftrightarrow } &{} \text {if} \vdash _{\mathcal {B}} a~ \text {then}~ \vdash _{\mathcal {B}} b.\\ \text {Where} ~p \not = b,&{} {\mathfrak {S}}_{\mathcal {B}}\, p &{} {\Leftrightarrow } &{} a \vdash _{\mathcal {B}} p.\\ \text {Where}~ P ~\text {is nonempty,}&{} P\,{\mathfrak {S}}_{\mathcal {B}}\,q &{} {\Leftrightarrow } &{} {\mathfrak {S}}_{\mathcal {C}}\,q~ \text {for every}~ {\mathcal {C}} \supseteq {\mathcal {B}}~ \text {such that} \\ &{}&{}&{} {\mathfrak {S}}_{\mathcal {C}}\,p~ \text {for every}~ p~ \text {in}~ P. \end{array} \end{aligned}$$

(The “if—then” in the first clause is to be understood materially, so that, whenever $\not \vdash _{\mathcal B} a$, vacuously $\mathrel {{\mathfrak S}_{\mathcal B}} b$.)

This $\mathfrak S$ satisfies (E) by definition. To see that it satisfies (R), note first that for any $\mathcal B$ and any $p$,

• (i) if $a \vdash _{\mathcal B} b$ then $\mathrel {{\mathfrak S}_{\mathcal B}} p$, and

• (ii) if $\mathrel {{\mathfrak S}_{\mathcal B}} p$ and $\vdash _{\mathcal B} a$ then $\vdash _{\mathcal B} p$.

(i) holds by definition where $p \not = b$, and, where $p = b$, in virtue of the fact that $\vdash _{\mathcal B} b$ if $a \vdash _{\mathcal B} b$ and $\vdash _{\mathcal B} a$. (ii), similarly, holds by definition where $p=b$, and where $p \not = b$ in virtue of the fact that $\vdash _{\mathcal B} p$ if $a \vdash _{\mathcal B} p$ and $\vdash _{\mathcal B} a$.

Now suppose that $\mathbf P \mathbin {\Rightarrow } q \in \mathcal B$, and moreover that $R \mathrel {{\mathfrak S}_{\mathcal B}} p$ for every $R \mathbin {\Rightarrow } p$ in $\mathbf P$. For any such $R \mathbin {\Rightarrow } p$, consider the basis

$$\begin{aligned} {\mathcal B}_{R,a} = \mathcal B \cup \{{}\mathbin {\Rightarrow } r \mathrel {|} r \in R\} \cup \{ \mathbin {\Rightarrow } a \}. \end{aligned}$$

For any $r$ in $R$ it holds that $\vdash _{{\mathcal B}_{R,a}} r$, whence $a \vdash _{{\mathcal B}_{R,a}} r$, whence $\mathrel {{\mathfrak S}_{{\mathcal B}_{R,a}}} r$ by (i). Since by hypothesis $R \mathrel {{\mathfrak S}_{\mathcal B}} p$ it follows that $\mathrel {{\mathfrak S}_{{\mathcal B}_{R,a}}} p$; and since moreover $\vdash _{{\mathcal B}_{R,a}} a$, by (ii) we may infer that $\vdash _{{\mathcal B}_{R,a}} p$, meaning that $a, R \vdash _{\mathcal B} p$.

Since this is true of every $R \mathbin {\Rightarrow } p$ in $\mathbf P$, and by hypothesis $\mathbf P \mathbin {\Rightarrow } q \in \mathcal B$, it follows that $a \vdash _{\mathcal B} q$, whence $\mathrel {{\mathfrak S}_{\mathcal B}} q$ by (i); this completes the verification of (R).

It remains to show that $\vdash$ is not included in $\mathfrak S$. Pick two formulas $c$ and $d$ distinct from $a$ and $b$, and consider the basis

$$\begin{aligned} \mathcal C = \{( {}\mathbin {\Rightarrow } a, {}\mathbin {\Rightarrow } b) \mathbin {\Rightarrow } c, (a \mathbin {\Rightarrow } c) \mathbin {\Rightarrow } d\}. \end{aligned}$$

By induction according to the definition of $\vdash$, it is easily verified that, in general, $P \vdash _{\mathcal C} q$ only if $q$ is a truth-functional consequence of $P \cup \{a \mathbin {\supset }(b \mathbin {\supset }c), (a \mathbin {\supset }c) \mathbin {\supset }d\}$. Therefore $\not \vdash _{\mathcal C} a$ and $a \not \vdash _{\mathcal C} d$, whence $\mathrel {{\mathfrak S}_{\mathcal C}} b$ but $\mathrel {{\not {\!\!\!\mathfrak S}}_{\mathcal C}} d$, whence $b \mathrel {{\not {\!\!\!\mathfrak S}}_{\mathcal C}} d$. Yet clearly $b \vdash _{\mathcal C} d$ since $b,a \vdash _{\mathcal C} c$; hence $\vdash$ is not a subrelation of $\mathfrak S$.

Thus, since $\vdash$ is the smallest $\mathfrak X$ satisfying (R), (E), and (M), it follows that, as claimed on page 319, $\mathfrak S$ is not a superrelation of any relation satisfying these conditions.