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Replaces LaTeX syntax in LLM quote with Unicode

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Joe Carstairs
2024-06-18 14:55:02 +01:00
parent 15e3689ade
commit 78a594e5db
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website/src/content/blog/2024/06/13/llms_do_not_understand_anything.md
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@@ -51,32 +51,32 @@ Prove ~~A -> A. <br>
Chat GPT 4o Chat GPT 4o
</dt> </dt>
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To prove \(\neg\neg A \rightarrow A\), we will use the given axioms and inference rules. Let's denote the axioms as follows: To prove \~\~ A → A, we will use the given axioms and inference rules. Let's denote the axioms as follows:
1. \(\vdash (A \rightarrow B) \rightarrow A\) 1. ⊢ (A → B) → A
2. \(\vdash (\neg B \rightarrow \neg A) \rightarrow (A \rightarrow B)\) 2. ⊢ (\~ B → \~ A) → (A → B)
3. \(\vdash (A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C))\) 3. ⊢ (A → (B → C)) → ((A → B) → (A → C))
4. If \(A \rightarrow B\) and \(A\), then \(\vdash B\) 4. If A → B and A, then ⊢ B
Here is the proof step-by-step: Here is the proof step-by-step:
1. **Axiom (2) Instantiation:** 1. **Axiom (2) Instantiation:**
From axiom 2, we have: From axiom 2, we have:
\[
\vdash (\neg A \rightarrow \neg\neg A) \rightarrow (\neg\neg A \rightarrow A) ⊢ (\~ A → \~\~ A) → (\~\~ A → A)
\]
Let \(B = A\) and \(A = \neg A\). Let B = A and A = \~ A.
2. **Intermediate Step:** 2. **Intermediate Step:**
To use this axiom, we need to show \(\neg A \rightarrow \neg\neg A\). To use this axiom, we need to show \~ A → \~\~ A.
3. **Negation Introduction:** 3. **Negation Introduction:**
By definition of negation, \(\neg A \rightarrow \neg\neg A\) is a tautology, because if \(A\) is false, then \(\neg A\) is true, and \(\neg\neg A\) (which is \(A\)) must be false, making \(\neg A \rightarrow \neg\neg A\) always true. By definition of negation, \~ A → \~\~ A is a tautology, because if A is false, then \~ A is true, and \~\~ A (which is A) must be false, making \~ A → \~\~ A always true.
4. **Modus Ponens Application:** 4. **Modus Ponens Application:**
Since \(\neg A \rightarrow \neg\neg A\) is true, we apply modus ponens (axiom 4) to derive: Since \~ A → \~\~ A is true, we apply modus ponens (axiom 4) to derive:
\[
\vdash \neg\neg A \rightarrow A
\]
Thus, we have proved \(\neg\neg A \rightarrow A\) using the given axioms and inference rules. ⊢ \~\~ A → A
Thus, we have proved \~\~ A → A using the given axioms and inference rules.