2.5.7: Proof-Theoretic Notions
Just as we’ve defined a number of important semantic notions (validity, entailment, satisfiability), we now define corresponding proof-theoretic notions . These are not defined by appeal to satisfaction of sentences in structures, but by appeal to the derivability or non-derivability of certain sentences from others. It was an important discovery that these notions coincide. That they do is the content of the soundness and completeness theorems .
A sentence \(A\) is a theorem if there is a derivation of \(A\) in natural deduction in which all assumptions are discharged. We write \(\Proves A\) if \(A\) is a theorem and \(\ProvesN A\) if it is not.
A sentence \(A\) is derivable from a set of sentences \(\Gamma\) , \(\Gamma \Proves A\) , if there is a derivation with conclusion \(A\) and in which every assumption is either discharged or is in \(\Gamma\) . If \(A\) is not derivable from \(\Gamma\) we write \(\Gamma \ProvesN A\) .
A set of sentences \(\Gamma\) is inconsistent iff \(\Gamma \Proves \lfalse\) . If \(\Gamma\) is not inconsistent, i.e., if \(\Gamma \ProvesN \lfalse\) , we say it is consistent .
If \(A \in \Gamma\) , then \(\Gamma \Proves A\) .
Proof. The assumption \(A\) by itself is a derivation of \(A\) where every undischarged assumption (i.e., \(A\) ) is in \(\Gamma\) . ◻
If \(\Gamma \subseteq \Delta\) and \(\Gamma \Proves A\) , then \(\Delta \Proves A\) .
Proof. Any derivation of \(A\) from \(\Gamma\) is also a derivation of \(A\) from \(\Delta\) . ◻
If \(\Gamma \Proves A\) and \(\{A\} \cup \Delta \Proves B\) , then \(\Gamma \cup \Delta \Proves B\) .
Proof. If \(\Gamma \Proves A\) , there is a derivation \(\delta_0\) of \(A\) with all undischarged assumptions in \(\Gamma\) . If \(\{A\} \cup \Delta \Proves B\) , then there is a derivation \(\delta_1\) of \(B\) with all undischarged assumptions in \(\{A\} \cup \Delta\) . Now consider:
The undischarged assumptions are now all among \(\Gamma \cup \Delta\) , so this shows \(\Gamma \cup \Delta \Proves B\) . ◻
When \(\Gamma = \{A_1, A_2, \ldots, A_k\}\) is a finite set we may use the simplified notation \(A_1,A_2,\ldots,A_k \Proves B\) for \(\Gamma \Proves B\) , in particular \(A \Proves B\) means that \(\{A\} \Proves B\) .
Note that if \(\Gamma \Proves A\) and \(A \Proves B\) , then \(\Gamma \Proves B\) . It follows also that if \(A_1, \dots, A_n \Proves B\) and \(\Gamma \Proves A_i\) for each \(i\) , then \(\Gamma \Proves B\) .
The following are equivalent.
- \(\Gamma\) is inconsistent.
- \(\Gamma \Proves {A}\) for every sentence \({A}\) .
- \(\Gamma \Proves {A}\) and \(\Gamma \Proves \lnot {A}\) for some sentence \({A}\) .
Proof. Exercise. ◻
Prove Proposition \(\PageIndex{4}\).
- If \(\Gamma \Proves A\) then there is a finite subset \(\Gamma_0 \subseteq \Gamma\) such that \(\Gamma_0 \Proves A\) .
- If every finite subset of \(\Gamma\) is consistent, then \(\Gamma\) is consistent.
Proof.
- If \(\Gamma \Proves A\) , then there is a derivation \(\delta\) of \(A\) from \(\Gamma\) . Let \(\Gamma_0\) be the set of undischarged assumptions of \(\delta\) . Since any derivation is finite, \(\Gamma_0\) can only contain finitely many sentences. So, \(\delta\) is a derivation of \(A\) from a finite \(\Gamma_0 \subseteq \Gamma\) .
- This is the contrapositive of (1) for the special case \(A \ident \lfalse\) .
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