joeac/joeac.net
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects 1 Releases Wiki Activity

Replaces LaTeX syntax in LLM quote with Unicode

Browse Source
This commit is contained in:
Joe Carstairs
2024-06-18 14:55:02 +01:00
parent 15e3689ade
commit 78a594e5db
1 changed files with 16 additions and 16 deletions
Download Patch File Download Diff File Expand all files Collapse all files

32
website/src/content/blog/2024/06/13/llms_do_not_understand_anything.md
View File

@@ -51,32 +51,32 @@ Prove ~~A -> A. <br>
Chat GPT 4o
</dt>
<dd>
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. \(\vdash (A \rightarrow B) \rightarrow A\)
2. \(\vdash (\neg B \rightarrow \neg A) \rightarrow (A \rightarrow B)\)
3. \(\vdash (A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C))\)
4. If \(A \rightarrow B\) and \(A\), then \(\vdash B\)
1. ⊢ (A → B) → A
2. ⊢ (\~ B → \~ A) → (A → B)
3. ⊢ (A → (B → C)) → ((A → B) → (A → C))
4. If A → B and A, then ⊢ B
Here is the proof step-by-step:
1. **Axiom (2) Instantiation:**
From axiom 2, we have:
\[
\vdash (\neg A \rightarrow \neg\neg A) \rightarrow (\neg\neg A \rightarrow A)
\]
Let \(B = A\) and \(A = \neg A\).
⊢ (\~ A → \~\~ A) → (\~\~ A → A)
Let B = A and A = \~ A.
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:**
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:**
Since \(\neg A \rightarrow \neg\neg A\) is true, we apply modus ponens (axiom 4) to derive:
\[
\vdash \neg\neg A \rightarrow A
\]
Since \~ A → \~\~ A is true, we apply modus ponens (axiom 4) to derive:
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.