2.5.8: 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. Let the derivation of \(A\) from \(\Gamma\) be \(\delta_1\) and the derivation of \(\lfalse\) from \(\Gamma \cup \{A\}\) be \(\delta_2\) . We can then derive:
In the new derivation, the assumption \(A\) is discharged, so it is a derivation from \(\Gamma\) . ◻
\(\Gamma \Proves A\) iff \(\Gamma \cup \{\lnot A\}\) is inconsistent.
Proof. First suppose \(\Gamma \Proves A\) , i.e., there is a derivation \(\delta_0\) of \(A\) from undischarged assumptions \(\Gamma\) . We obtain a derivation of \(\lfalse\) from \(\Gamma \cup \{\lnot A\}\) as follows:
Now assume \(\Gamma \cup \{\lnot A\}\) is inconsistent, and let \(\delta_1\) be the corresponding derivation of \(\lfalse\) from undischarged assumptions in \(\Gamma \cup \{\lnot A\}\) . We obtain a derivation of \(A\) from \(\Gamma\) alone by using \(\FalseCl\) :
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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 \(\delta\) of \(A\) from \(\Gamma\) . Consider this simple application of the \(\Elim{\lnot}\) rule:
Since \(\lnot A \in \Gamma\) , all undischarged assumptions are in \(\Gamma\) , this shows that \(\Gamma \Proves \lfalse\) . ◻
If \(\Gamma \cup \{A\}\) and \(\Gamma \cup \{\lnot A\}\) are both inconsistent, then \(\Gamma\) is inconsistent.
Proof. There are derivations \(\delta_1\) and \(\delta_2\) of \(\lfalse\) from \(\Gamma \cup \{ A \}\) and \(\lfalse\) from \(\Gamma \cup \{ \lnot A \}\) , respectively. We can then derive
Since the assumptions \(A\) and \(\lnot A\) are discharged, this is a derivation of \(\lfalse\) from \(\Gamma\) alone. Hence \(\Gamma\) is inconsistent. ◻