2.4.9: Derivability and Consistency
We will now establish a number of properties of the derivability relation. They are independently interesting, but each will play a role in the proof of the completeness theorem.
If \(\Gamma \Proves A\) and \(\Gamma \cup \{A\}\) is inconsistent, then \(\Gamma\) is inconsistent.
Proof. There are finite \(\Gamma_0\) and \(\Gamma_1 \subseteq \Gamma\) such that \(\Log{LK}\) derives \(\Gamma_0 \Sequent A\) and \(A, \Gamma_1 \Sequent \quad\) . Let the \(\Log{LK}\) -derivation of \(\Gamma_0 \Sequent A\) be \(\pi_0\) and the \(\Log{LK}\) -derivation of \(\Gamma_1, A \Sequent \quad\) be \(\pi_1\) . We can then derive
Since \(\Gamma_0 \subseteq \Gamma\) and \(\Gamma_1 \subseteq \Gamma\) , \(\Gamma_0 \cup \Gamma_1 \subseteq \Gamma\) , hence \(\Gamma\) is inconsistent. ◻
\(\Gamma \Proves A\) iff \(\Gamma \cup \{\lnot A\}\) is inconsistent.
Proof. First suppose \(\Gamma \Proves A\) , i.e., there is a derivation \(\pi_0\) of \(\Gamma \Sequent A\) . By adding a \(\LeftR{\lnot}\) rule, we obtain a derivation of \(\lnot A, \Gamma \Sequent \quad\) , i.e., \(\Gamma \cup \{\lnot A\}\) is inconsistent.
If \(\Gamma \cup \{\lnot A\}\) is inconsistent, there is a derivation \(\pi_1\) of \(\lnot A, \Gamma \Sequent \quad\) . The following is a derivation of \(\Gamma \Sequent A\) :
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Prove that \(\Gamma \Proves \lnot A\) iff \(\Gamma \cup \{A\}\) is inconsistent.
If \(\Gamma \Proves A\) and \(\lnot A \in \Gamma\) , then \(\Gamma\) is inconsistent.
Proof. Suppose \(\Gamma \Proves A\) and \(\lnot A \in \Gamma\) . Then there is a derivation \(\pi\) of a sequent \(\Gamma_0 \Sequent A\) . The sequent \(\lnot A, \Gamma_0 \Sequent \quad\) is also derivable:
Since \(\lnot A \in \Gamma\) and \(\Gamma_0 \subseteq \Gamma\) , this shows that \(\Gamma\) is inconsistent. ◻
If \(\Gamma \cup \{A\}\) and \(\Gamma \cup \{\lnot A\}\) are both inconsistent, then \(\Gamma\) is inconsistent.
Proof. There are finite sets \(\Gamma_0 \subseteq \Gamma\) and \(\Gamma_1 \subseteq \Gamma\) and \(\Log{LK}\) -derivations \(\pi_0\) and \(\pi_1\) of \(A, \Gamma_0 \Sequent \quad\) and \(\lnot A, \Gamma_1 \Sequent \quad\) , respectively. We can then derive
Since \(\Gamma_0 \subseteq \Gamma\) and \(\Gamma_1 \subseteq \Gamma\) , \(\Gamma_0 \cup \Gamma_1 \subseteq \Gamma\) . Hence \(\Gamma\) is inconsistent. ◻