A set of sentences is consistent when some row makes every one of them true. When no row does, the set is inconsistent, and the smallest subset that already fails is the place to look. That subset says that something must go. It does not say what.
Core ideas
An argument has a conclusion to test. A set of beliefs has none, and the question changes from whether something follows to whether the sentences can live together.
Take the table over the simple sentences in the set. A row that makes every member true is a model. A set with at least one model is consistent; a set with none is inconsistent. Counting the models says how much the set leaves open.
Put a conclusion's denial beside the premises. If the result has no model, no row makes the premises true and the conclusion false, which is validity. Testing an argument and testing a belief set are the same table read two ways.
Remove sentences until the rest can all be true. A subset that cannot all be true, and becomes possible when any member is removed, is a minimal inconsistent subset. It marks where the trouble is, and nothing in it marks which member is the false one.
The material
The test is the table of “Validity by Truth Table”, run on a set instead of an argument.
A set of sentences is consistent when at least one row of its table makes every member true. Such a row is a model of the set. A set with no model is inconsistent.
Start with the smallest interesting case. The set contains p → q, p and ¬q. Read the three together: if the first two hold, q must hold, and the third denies it. This is modus ponens with its conclusion turned into a premise, and the table shows that no row survives.
Every row has a false cell, so there is no model and the set is inconsistent. The lab reports this as zero rows of four. Note what kind of fact this is. No pair of these sentences contradicts another: p → q with p is fine, p with ¬q is fine, and p → q with ¬q is fine. The trouble appears only when all three are held at once.
A subset that cannot all be true, and that becomes possible when any one member is removed, is a minimal inconsistent subset. A set may have several. The lab reports the smallest, and among those of equal size the one that comes first by sentence number.
Sentences outside the minimal subset are innocent of this particular failure. Add q → r to the set above and the smallest failing subset is still the first three. Added sentences cannot repair an inconsistency, because every row that fails already fails.
Now the part that matters for philosophy. An inconsistent set tells you that at least one member is false. It does not tell you which. In the four-sentence set p → q, q → r, p, ¬r the lab lets you give up each sentence in turn, and each time exactly one model appears:
All four repairs work, and they describe four different worlds. Which to choose is a question about the sentences themselves: how much you trust each, what each costs to lose, what else depends on it. The table is neutral. That neutrality is why a skeptic's valid argument forces a choice and cannot make it.
Consistency is also weaker than truth. The set p ∨ q, ¬p, q → r is consistent, with one model, p false and q and r true. That means the three sentences could all be true. Whether they are is again a question about the world, which no lab here answers.
Interactive
Every model, and every minimal inconsistent subset
Method
Four steps. The last is a judgement, and the lab stops before it.
Use one letter per simple claim and the same letter every time the claim recurs. A letter used for two different claims manufactures an inconsistency that was never there.
Find a row that makes every sentence true. If you find one, the set is consistent and you can stop: that one row is the whole proof.
Give up each sentence in turn, or try subsets by size. The smallest subset that fails is where the incompatibility lives.
Any member of a minimal subset can be dropped to restore consistency. Pick by which you have most reason to doubt, and say what the choice costs.
Let p be “the alarm was set”, q “the alarm rang” and r “she woke”. Four beliefs: if it was set it rang; if it rang she woke; it was set; she did not wake.
Chain the first three: set, so rang, so woke. The fourth denies the last link. No smaller subset fails, because giving up any one of the four leaves a possible world, and the table above lists all four. If the alarm was not set the story is one thing; if it rang without waking her, another; if she did wake, a third. The lab has shown that the beliefs cannot stand together and has not said which is false.
Given a set of up to four sentences over three letters, either write a row that makes all of them true or name the smallest subset that cannot all be true, and say what that subset does and does not tell you about which belief to give up.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.