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