2.4.9: Derivability and Consistency

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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.

Proposition \(\PageIndex{1}\)

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. ◻

Proposition \(\PageIndex{2}\)

\(\Gamma \Proves A\) 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 \(\Gamma \Sequent A\):

◻

Problem \(\PageIndex{1}\)

Prove that \(\Gamma \Proves \lnot A\) iff \(\Gamma \cup \{A\}\) is inconsistent.

Proposition \(\PageIndex{3}\)

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. ◻

Proposition \(\PageIndex{4}\)

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. ◻