Lesson · Knowledge and Evidence · Justification

The Regress of Justification

If every justified belief needs a justified supporter, the chain of support must end in a foundation, loop back on itself, or run on forever. The set of claims that rules out all three is inconsistent, its smallest inconsistent subset is the whole set, and each single deletion gives one named position.

Open the interactive lesson Practice

some belief is justified: j j needs a justified supporter: j → n n → circle, foundation or infinite chain deny each of the three: inconsistent

Six claims, no model

Core ideas

What this lesson establishes

The regress is an old problem, and the earlier course gave us a tool for it: a set of claims is consistent when some assignment makes them all true.

The set is inconsistent

The six sentences have no model. Every assignment makes at least one false, so something has to give, and the question becomes which.

A minimal inconsistent subset is what is really at issue

It is a set that cannot all be true and in which removing any one sentence restores consistency. Here it is the whole set, so every sentence is load-bearing for the trouble.

Each deletion is a position

Delete one of the three denials and a way of ending the regress is permitted. Delete the first sentence and you have the skeptic.

The material

Agrippa’s trilemma as a set of claims

Six sentences, one inconsistency, and four ways out.

Suppose you believe the tap is dripping because you heard it. What justifies hearing as a source? That it has been reliable. What justifies that? A track record of checks, each of which was itself a belief. Asked for the support of each supporter, you can answer in only a few ways, and the old skeptical argument says none of them is satisfactory.

Write it as six claims over five letters. j says some belief is justified. n says every justified belief needs a justified supporter. And c, f and i say that, respectively, a circle of supporters, a foundation that needs no further support, and an infinite chain of supporters are acceptable ways for the chain to end.

1. j some belief is justified 2. j → n so it needs a justified supporter 3. n → (c ∨ f ∨ i) the chain is a circle, a foundation or infinite 4. ¬c circles are not acceptable 5. ¬f unsupported foundations are not 6. ¬i infinite chains are not

The lab finds no assignment that makes all six true, and reports the smallest inconsistent subset as all six sentences. That matters. A minimal inconsistent subset is the part of the set that is doing the damage, and here no part of it can be spared: remove any one sentence and a model appears.

Four positions, four deletions

Foundationalism deletes ¬f. Some beliefs justify others without needing a justifier of their own, and the lab’s model has f true and the chain stopping there.
Coherentism deletes ¬c. Beliefs support one another in a web, and the objection that this is a circle is denied as an objection.
Infinitism deletes ¬i. Justification is never completed, only deepened as far as questions are asked.
Skepticism deletes j. If nothing is justified there is nothing for the regress to apply to, and the lab’s skeptic preset replaces j with ¬j.

The foundationalist version is one line of a truth table. Take the foundationalist set, with ¬f removed, and the lab finds a model in which j, n and f are true and c and i are false: a justified belief that needs a supporter, the supporter being a foundation. That the set is satisfiable shows only that the position is coherent as a set of sentences. Whether any belief is a foundation is the substantive question.

One limit should be stated. The lab sees the five letters as unanalysed claims. It does not look inside “every justified belief needs a supporter” to see a chain, so the structure of the regress is carried by the sentences, not computed from them. Everything the lab reports follows from the six sentences as written.

Interactive

Consistency and what must go

Every model, and every minimal inconsistent subset

Method

Reading the regress as a consistency problem

From a puzzle in words to a set the lab can search.

Write every commitment as a sentence

Include the denials. The trilemma is only a problem because the skeptic denies each ending, and a denial left unwritten is a premise hidden.

Check the whole set

Ask for a model. If the lab finds none, the set is inconsistent and something has to be deleted.

Read the minimal inconsistent subset

It names the sentences the trouble actually needs. If it is smaller than the set, the rest is innocent.

Delete one sentence at a time

For each deletion read the model the lab finds. Each model is a picture of one position: which of the endings it allows.

Name the cost of each position

Each position keeps the set consistent by permitting an ending the skeptic found unacceptable. Say what the objection was and how the position answers it.

Positions as deletions

Score 0 / 0

Choose the answer you can justify.

Dropping the denial of foundations

Compare the whole set with the foundationalist set.

all six: 0 models, Inconsistent smallest MIS: the whole set drop ¬f (5): consistent, j n f true; c i false drop ¬c (4): consistent, c true drop j (1): consistent, nothing justified

Dropping a different sentence gives a different position, and the lab’s model for each says which ending the position leans on. The positions are not refuted by this. Each one is what the set looks like when one of its denials is given up.

Common mistakes

Treating “circular” and “infinite” as the same complaintThe set gives them different letters. A circle complains that a claim ends up supporting itself; an infinite chain complains that no one can ever finish giving the support. Deleting ¬c yields a model in which c is true and i false, deleting ¬i the other way round, and the positions that result, coherentism and infinitism, answer different objections.

Thinking the inconsistency refutes epistemologyAn inconsistent set says that something in it has to be given up, not which thing. The skeptic gives up j, the others give up a denial. The argument has no force until a reason is given for treating the six sentences as fixed, and each of the denials can be argued against.

Expecting the lab to find which position is rightThe lab finds models, and every deletion has one. Choosing among them depends on how costly each ending is to accept, and that is the judgement the lesson leaves open on purpose.

Check your understanding

✓

Finish when you can name the position for each deletion.

You should be able to write the trilemma as a set of sentences, state its smallest inconsistent subset, and for each single deletion say which position results and which letter the lab’s model makes true.

Note: The same procedure works for any philosophical puzzle that comes as a list of individually appealing claims. Write them down, ask whether they can all be true, and see which deletion you can live with.