2.4.8: 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 sequents. It was an important discovery that these notions coincide. That they do is the content of the soundness and completeness theorem .
A sentence \(A\) is a theorem if there is a derivation in \(\Log{LK}\) of the sequent \(\quad \Sequent A\) . 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 \(\Gamma\) , \(\Gamma \Proves A\) , iff there is a finite subset \(\Gamma_0 \subseteq \Gamma\) and a sequence \(\Gamma_0'\) of the sentences in \(\Gamma_0\) such that \(\Log{LK}\) derives \(\Gamma_0' \Sequent A\) . If \(A\) is not derivable from \(\Gamma\) we write \(\Gamma \ProvesN A\) .
Because of the contraction, weakening, and exchange rules, the order and number of sentences in \(\Gamma_0'\) does not matter: if a sequent \(\Gamma_0' \Sequent A\) is derivable, then so is \(\Gamma_0'' \Sequent A\) for any \(\Gamma_0''\) that contains the same sentences as \(\Gamma_0'\) . For instance, if \(\Gamma_0 = \{B, C\}\) then both \(\Gamma_0' = \tuple{B, B, C}\) and \(\Gamma_0'' = \tuple{C, C, B}\) are sequences containing just the sentences in \(\Gamma_0\) . If a sequent containing one is derivable, so is the other, e.g.:
From now on we’ll say that if \(\Gamma_0\) is a finite set of sentences then \(\Gamma_0 \Sequent A\) is any sequent where the antecedent is a sequence of sentences in \(\Gamma_0\) and tacitly include contractions, exchanges, and weakenings if necessary.
A set of sentences \(\Gamma\) is inconsistent iff there is a finite subset \(\Gamma_0 \subseteq \Gamma\) such that \(\Log{LK}\) derives \(\Gamma_0 \Sequent \quad\) . If \(\Gamma\) is not inconsistent, i.e., if for every finite \(\Gamma_0 \subseteq \Gamma\) , \(\Log{LK}\) does not derive \(\Gamma_0 \Sequent \quad\) , we say it is consistent .
If \(A \in \Gamma\) , then \(\Gamma \Proves A\) .
Proof. The initial sequent \(A \Sequent A\) is derivable, and \(\{A\} \subseteq \Gamma\) . ◻
If \(\Gamma \subseteq \Delta\) and \(\Gamma \Proves A\) , then \(\Delta \Proves A\) .
Proof. Suppose \(\Gamma \Proves A\) , i.e., there is a finite \(\Gamma_0 \subseteq \Gamma\) such that \(\Gamma_0 \Sequent A\) is derivable. Since \(\Gamma \subseteq \Delta\) , then \(\Gamma_0\) is also a finite subset of \(\Delta\) . The derivation of \(\Gamma_0 \Sequent A\) thus also shows \(\Delta \Proves A\) . ◻
If \(\Gamma \Proves A\) and \(\{A\} \cup \Delta \Proves B\) , then \(\Gamma \cup \Delta \Proves B\) .
Proof. If \(\Gamma \Proves A\) , there is a finite \(\Gamma_0 \subseteq \Gamma\) and a derivation \(\pi_0\) of \(\Gamma_0 \Sequent A\) . If \(\{A\} \cup \Delta \Proves B\) , then for some finite subset \(\Delta_0 \subseteq \Delta\) , there is a derivation \(\pi_1\) of \(A, \Delta_0 \Sequent B\) . Consider the following derivation:
Since \(\Gamma_0 \cup \Delta_0 \subseteq \Gamma \cup \Delta\) , this shows \(\Gamma \cup \Delta \Proves B\) . ◻
Note that this means that in particular 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 \(\Gamma \Proves B\) .
\(\Gamma\) is inconsistent iff \(\Gamma \Proves {A}\) for every 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 finite subset \(\Gamma_0 \subseteq \Gamma\) such that the sequent \(\Gamma_0 \Sequent A\) has a derivation. Consequently, \(\Gamma_0 \Proves A\) .
- If \(\Gamma\) is inconsistent, there is a finite subset \(\Gamma_0 \subseteq \Gamma\) such that \(\Log{LK}\) derives \(\Gamma_0 \Sequent \quad\) . But then \(\Gamma_0\) is a finite subset of \(\Gamma\) that is inconsistent.
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