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Chapter 18: Problem 4

Prove that \(\Gamma \vdash \neg \varphi\) iff \(\Gamma \cup\\{\varphi\\}\) is inconsistent.

Short Answer

Expert verified
The proof has two parts. For the first part, if \(螕 \vdash 卢蠁\), then it implies that \(螕 \cup \{蠁\}\) is inconsistent because if \(螕 \cup \{蠁\}\) was not inconsistent, i.e. no contradiction can be derived, it would mean that 蠁 and 卢蠁 (which is implied by 螕) could both be true, which is a contradiction. For the second part, if \(螕 \cup \{蠁\}\) is inconsistent, it implies that \(螕 \vdash 卢蠁\). Because by being inconsistent, it means a contradiction can be derived from 螕 and 蠁, so 卢蠁 must be an entailment of 螕.

Step by step solution

01

Proving First Implication

The first part of the 'iff' statement is '\(螕 \vdash 卢蠁\) implies \(螕 \cup \{蠁\}\) is inconsistent'. This means that if 螕 entails 卢蠁, then adding 蠁 to 螕 makes it inconsistent. Start by assuming that \(螕 \vdash 卢蠁\). This means that from 螕, it is possible to derive 卢蠁. Next, suppose for contradiction that \(螕 \cup \{蠁\}\) is not inconsistent, i.e. no contradiction can be derived from it. But this would mean that 螕 and 蠁 can simultaneously be true, contradicting the initial assumption that \(螕 \vdash 卢蠁\) (since if 螕 and 蠁 are true then 卢蠁 would be false). Therefore, the assumption must be incorrect and so \(螕 \cup \{蠁\}\) has to be inconsistent.
02

Proving Second Implication

The second part of the 'iff' statement means that if \(螕 \cup \{蠁\}\) is inconsistent, then 螕 entails 卢蠁. This implies that every model that makes 螕 true makes 蠁 false. Start by assuming that \(螕 \cup \{蠁\}\) is inconsistent. This means that there is a contradiction derived from 螕 and 蠁. Since 卢蠁 is a contradiction of 蠁 and no contradiction can be derived in an inconsistent set, this means 螕 entails 卢蠁. If not, then there would exist a case for which 螕 is true and 蠁 is true, which would contradict our assumption that \(螕 \cup \{蠁\}\) is inconsistent. This completes the proof for the second implication.
03

Wrapping up

Since it has been shown that \(螕 \vdash 卢蠁\) if and only if \(螕 \cup \{蠁\}\) is inconsistent, it concludes the proof for the initial task 'Prove that \(螕 \vdash 卢蠁\) iff \(螕 \cup \{蠁\}\) is inconsistent'. It's a well-structured proof that applies the definition of logical entailment, contradiction, and consistency.

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Most popular questions from this chapter

Give derivations of the following: 1\. \(\neg(\varphi \rightarrow \psi) \rightarrow(\varphi \wedge \neg \psi)\) 2\. \((\varphi \rightarrow \chi) \vee(\psi \rightarrow \chi)\) from the assumption \((\varphi \wedge \psi) \rightarrow \chi\) 3\. \(\neg \neg \varphi \rightarrow \varphi\) 4\. \(\neg \varphi \rightarrow \neg \psi\) from the assumption \(\psi \rightarrow \varphi\) 5\. \(\neg \varphi\) from the assumption \((\varphi \rightarrow \neg \varphi)\) 6\. \(\varphi\) from the assumptions \(\psi \rightarrow \varphi\) and \(\neg \psi \rightarrow \varphi\)

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