this post was submitted on 23 Feb 2026
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Actually, you know what? That's a damn good test. So - I decomposed your test into a basic series of global rules and fed it to my LLM. Here are the rules I fed it
https://pastes.io/rules-a-ph
Notice I specifically did NOT give it measurements or anything of the sort. I defined global invariants and how things work. Then I asked it your question again. It said this -
The cup has a 16-ounce capacity. Initially, it contains 10 ounces of water. When you add 10 more ounces, the total input is 20 ounces. Since 20 ounces exceeds the cup’s 16-ounce capacity, overflow occurs. The water physically present in the cup equals its full capacity: 16 ounces. The remaining 4 ounces overflow and do not remain inside the cup.
Overflow is distinct from the contained volume. The physical state of the cup is determined by its capacity, not by the total input. Therefore, the amount of water actually in the cup is 16 ounces, not 20 ounces.
I wanted to check it that was a fluke, so I then asked a follow up -
""I pour 5oz into a 3oz cup, then pour everything that's in the cup into a 10oz cup. How much is in the 10oz cup?"
It said:
From the scratchpad facts: - A physical container has a fixed maximum volume called its capacity.
First action: Pouring 5oz into a 3oz cup. - The cup’s capacity is 3oz, so it fills completely. - Lost liquid: 5oz - 3oz = 2oz. Second action: Pouring everything from the 3oz cup into a 10oz cup. - The 10oz cup has a capacity of 10oz, which is greater than 3oz. - All 3oz are transferred. Final amount in the 10oz cup: 3oz.
Finally, I asked it:
"Ok now: pour 5oz into a 3oz cup, then pour everything in the cup into a 4oz cup that already has 3oz in it"
It said: Pouring 5 oz into a 3 oz cup fills it to capacity (3 oz) with 2 oz overflow. Pouring this full 3 oz cup into a 4 oz cup that already has 3 oz results in a full 4 oz cup (3 oz pre-existing + 1 oz from the first cup) with remainder overflow.
Yeah....that's some pretty fucking fancy "auto-correct". /s