A thousand planks are replaced one by one. Replacing one plank seems to preserve identity, and a thousand times over that gives an all-new ship that is the original. The lab runs the chain under the classical, cutoff and degree treatments, and the question becomes which premise or which logic to give up.
Core ideas
The five modal lessons asked what is true at the worlds a world can see. The next three ask what stays the same through time, and the oldest case is a ship. The ship of Theseus is a sorites in disguise. Treating it as one lets the lab count the steps and show what each way out of the chain commits you to.
Let F(k) say that the ship with k original planks, repaired plank by plank, is the original ship. “Replacing one plank preserves identity” is F(k) → F(k − 1), and there is one such conditional for each plank.
From F(1000) and every conditional, modus ponens a thousand times gives F(0): the ship with no original planks is the original. If the conclusion is unacceptable, a premise or the logic must go.
The classical treatment keeps every conditional and accepts the conclusion. The cutoff treatment makes one conditional false. The degree treatment makes each slightly less than true, and the conclusion has value 0.
The material
The conditionals, the lab’s counts, and what is left to say about the reassembled ship.
Begin with the puzzle in its standard form. A wooden ship has a thousand planks. Each year one is replaced. After a thousand years none of the original planks remains, and it seems to have been the same ship throughout, since at no year did anyone say that a different ship had appeared. But a ship with no original plank is not the original ship, at least not on the face of it.
Write the first claim as a chain. The starting ship has all its planks, F(1000). The tolerance claim, one plank does not matter, is the family F(1000) → F(999), F(999) → F(998) and so on down to F(1) → F(0). The conclusion is F(0). This is the structure of a sorites, the same chain as in “Slippery Slopes and Small Differences” in Ethics and the Arithmetic of Welfare, and the lab counts it: 1000 steps, 1000 conditionals.
Under the classical treatment every conditional is true, and the lab prints “all 1000 true” and the conclusion True. The ship with no original plank is the original ship. Accepting that is the first exit, and it is not as silly as it sounds: it says that what makes a ship the same is continuity of repair, not material.
Under the cutoff treatment one conditional is false. With the cutoff at 500 it is F(500) → F(499): the ship with 500 original planks is the original, and the one with 499 is not. The lab prints that one conditional is false, at 500, and the conclusion False. Nothing in the arithmetic chose 500. Any number from 1 to 1000 gives a consistent verdict, and a reader who picks one has made an input and not found a fact.
Under the degree treatment each conditional is almost true. Each loses one thousandth, so each has value 999/1000, and a thousand of them chained lose everything: the lab prints 0 as the value the chain guarantees for the conclusion. Identity comes in degrees, and the all-new ship is guaranteed none of it.
The standard puzzle adds a second ship. The old planks, as they were removed, were stored and later reassembled in their old order. The reassembled ship has every original plank, and was never repaired. Now there are two candidates for being Theseus’ ship, and they are two ships, so at most one is the original.
The lab says nothing about the reassembled ship directly, because its claim is not a step in the chain. What it computes is how much the repaired ship keeps. Classically it keeps the whole title, so the reassembled ship does not have it. With a cutoff at 500 the repaired ship loses the title at one replacement, and a reader who thinks material makes a ship the same will give it to the reassembled one. With degrees the repaired ship ends at 0, and the reassembled one has the only remaining claim. Which criterion should carry the weight, continuity of repair or the parts, is the philosophical question, and the lab counts only the first.
So the arithmetic does not find the original ship. It shows what a treatment delivers when you feed it a chain. The choice of treatment is the question.
Interactive
A chain of tolerance conditionals under four treatments of vagueness
Method
Five moves. The first two set up the chain and the rest read the treatments.
“Is the original ship” applied to the ship with all its planks is the starting case. Write it as F at the starting count.
One step changes the count by one, and the claim is that this does not change the answer. Write F(k) → F(k − 1).
The start minus the end. The lab prints it, and it is also the number of conditionals.
Classical keeps every conditional and accepts the end. Cutoff falsifies one. Degrees lower each, and the conclusion has value 0.
A reassembled or rival ship is not a step of the chain. Ask which criterion would give it the title and whether you accept that criterion.
The starting count is 1000, the end count 0, and the cutoff is 500.
Three treatments of one chain, and three different answers about the same ship. The arithmetic reported each one exactly; it did not say which treatment to use.
You should be able to state the tolerance conditional for the ship, give the number of steps, read the lab’s report under the classical, cutoff and degree treatments, and say which premise or which logic each one gives up. A verdict stated without the treatment it came from has not been computed.
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