Suppose each half of a brain is placed in a new body, and both survivors are continuous with the original. Continuity seems enough for survival, and identity is one-one, and the two claims cannot both hold. The consistency lab finds no model for the five sentences, shows that dropping any one restores a model, and so lists the exits.
Core ideas
The last lesson ended with a chain that runs in a line. Here the chain forks, and the method is the one “Consistency and Belief Sets” taught and “The Regress of Justification” in Knowledge and Evidence applied: write the claims as a set and ask for a model. Two later lessons in this course use it again.
The case is stipulated: after the division two people, A and B, are each psychologically continuous with the original person. What is in dispute is how to describe it, and the description is a set of sentences the lab can test.
Write ca for “A is continuous with me” and ia for “A is me”, and likewise cb and ib. The claims are ca, cb, ca → ia, cb → ib and ¬(ia ∧ ib). No row of the truth table makes all five true.
Drop one sentence and a model appears. Each deletion is a position that has been defended, and a reader can choose among them by what each costs.
The material
The case, the five sentences, the models that appear when one is dropped, and the exit that changes the question.
The case is a thought experiment, and it is stated so that nothing in it is impossible by the laws of logic. Each half of a person's brain can sustain a mind, and the two halves are removed and placed in two new bodies. Each of the two people who wake is psychologically continuous with the original in every way the previous lesson asked for: memories, intentions, character. Call them A and B.
Three things are then hard to deny together. A is continuous with the original, and B is, because that is how the case was built. Continuity is enough for survival: if someone were connected to me by the chain I would want, that someone is me. And identity is one-one: if A is me and B is me, then A is B, but A and B now lead two lives and have two different bodies. Written out, with the sufficiency claim split over the two survivors, these are five sentences.
The lab's consistency mode, with the rows of the truth table behind it, finds no row that makes all five true. By sentences 1 and 3, ia is true; by 2 and 4, ib is true; sentence 5 denies their conjunction. The smallest inconsistent subset it reports is all five, so every sentence is needed for the contradiction, and the drop menu shows what each is worth: remove any one and a model is found.
A set of sentences with no model is minimally inconsistent when removing any one sentence leaves a set that has a model. Every sentence of such a set is doing work, and each is a place where the set can be repaired.
Dropping sentence 1 or 2 denies the stipulation, and says only one survivor is truly continuous. That changes the case instead of answering it. Dropping sentence 3 or 4 is the more interesting move. A closest-continuer theory says continuity gives identity only when no rival is as close, so that in the case of a tie neither survivor is me. The lab's second preset writes this as continuity suffices only when unbranched, and the set then has a model. The theory adds that in a tie neither survivor is me, and its price is that my survival depends on whether someone else also exists: if only one half had taken, I would have survived, and with both I do not.
Dropping sentence 5 says both A and B are me. One version holds that there were always two people before the division, overlapping in one body, so that the one-one claim is true of people and false of the stages we counted as one. The cost is a count of persons that no one made before the case arose.
The exit that changes the question is Parfit's. Identity, he says, is the wrong thing to want. What matters in survival is the relation of continuity and connectedness, and both A and B have it in full. The third preset replaces sentences 3 and 4 by claims that continuity gives what matters, written ca → ma and cb → mb, and keeps the one-one claim, and the set has a model. Whether A is me is left without an answer that matters, and it is the loss of that answer, not of a person, that Parfit asks you to accept. Whether he is right to say that nothing is lost is the philosophical question, and the lab, which can only test the set, leaves it open.
Interactive
Every model, and every minimal inconsistent subset
Method
Five moves, in this order. The third is the one that is usually done by instinct.
Write what the thought experiment stipulates, and nothing that is merely natural to say. Here those are the two continuity claims.
Give the sufficiency of continuity and the one-one nature of identity their own lines. A principle left implicit is a premise nobody can reject.
If the lab finds none, the set is inconsistent. Read the smallest inconsistent subset: it names the sentences that must be present for the trouble.
Read the model that appears each time, and say which view of identity it is.
A deletion is a position only when you can say what it gives up. For each exit, name a case in which the cost shows.
Letters: ca and cb for continuity with A and B, ia and ib for being me.
Nothing in the table says which sentence to keep. Dropping sentence 3 leaves a model in which only B is me, and dropping sentence 4 one in which only A is, and the case was built so that nothing prefers either. A view that keeps both conditionals and yet wants a single survivor has to add a tie-break the five sentences do not contain.
You should be able to write the five sentences, run the check and report that no model exists, delete each sentence in turn and say which position the model that appears represents, and state what Parfit's exit gives up. A deletion named without the model it yields has not been checked.
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