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+title: LLMs do not understand anything
+description: >-
+  Save this for the next time someone tells you that LLMs 'understand' things.
+pubDate: 2024-06-13
+
+content: |
+  LLMs do not understand what they are talking about. They just don't. It's not
+  how they're built. They have a statistical model of language, not a semantic
+  one. Philosophical puzzles about whether silicon can be conscious
+  _do not arise_, because LLMs are _not even close to the right shape_ for having
+  _anything like_ 'understanding'.
+  
+  If you don't believe me, there are plenty of examples out there on the Internet,
+  but this is as good as any. It includes Chat GPT 4o explaining in detail why
+  \~\~A → \~A (which is a classical contradiction) is trivially true in classical
+  logic.
+  
+  It's even better given that I had explicitly asked it to explain why that
+  sentence implies the trivial logic, not why it is trivially true. And even had
+  the explanation not been complete garbage from beginning to end, it would only
+  have shown that the sentence was _true_, not that it was trivial.
+  
+  In other words, the output:
+  
+  - Attempts to prove a contradiction (unprompted!)
+  - Confuses the concepts 'truth' and 'triviality'
+  - Is irrelevant to the prompt
+  
+  In case you want it handy, I'll put the full conversation down below. The good
+  bit is at the end. The only edits I've made are to replace TeX syntax with
+  Unicode.
+  
+  ---
+  
+  <dl>
+  <dt>
+  Me
+  </dt>
+  <dd>
+  You have the following axioms. <br>
+  1. |- (A -> B) -> A <br>
+  2. |- (~B -> ~A) -> (A -> B) <br>
+  3. |- (A -> (B -> C)) -> ((A -> B) -> (A -> C)) <br>
+  4. A -> B, A |- B <br>
+  Prove ~~A -> A. <br>
+  </dd>
+  
+  <dt>
+  Chat GPT 4o
+  </dt>
+  <dd>
+  To prove ~~ A → A, we will use the given axioms and inference rules. Let's denote the axioms as follows:
+  
+  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:
+  
+     ⊢ (\~ A → \~\~ A) → (\~\~ A → A)
+  
+     Let B = A and A = \~ A.
+  
+  2. **Intermediate Step:**  
+     To use this axiom, we need to show \~ A → \~\~ A.
+  
+  3. **Negation Introduction:**  
+     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 \~ A → \~\~ A is true, we apply modus ponens (axiom 4) to derive:
+  
+     ⊢ \~\~ A → A
+  
+  
+  Thus, we have proved \~\~ A → A using the given axioms and inference rules.