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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
----
-
-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.