The liar sentence says of itself that it is not true. With the truth schema and the rule that every sentence is true or not true, it follows that it is true exactly when it is not. The truth table for p ↔ ¬p has no satisfying row, which shows that the premises cannot all hold and says nothing about which to give up. Tarski's answer is to forbid the sentence.
Core ideas
The barber could be left out of the village. The liar is a sentence we can write down, and that is what makes it harder.
Let L be the sentence “L is not true”. It is built by ordinary grammar, with a subject that names the sentence and a predicate that denies truth of it. “Not true” is used, rather than “false”, for a reason that the lesson explains.
That L says it is not true; that a sentence is true exactly when things are as it says; and that every sentence is either true or not true. With p for “L is true” they yield p ↔ ¬p, which no row of a truth table makes true.
“The liar is simply false” is a row of the table, and the table rejects it. What the table cannot do is choose among the premises, which is the work the exits are for.
The material
The argument, the table that shows it closes, and the exit that most of the literature starts from.
Take a sentence that talks about itself. Print the following in a box and let L be the sentence in the box: “the sentence in this box is not true”. Nothing in the grammar is odd, and a sentence can refer to itself by being the sentence in a named place, as a label on a jar can refer to the jar. Ask whether L is true.
The first premise is a matter of how L was built. The second is the truth schema: “snow is white” is true exactly when snow is white, and the same for every sentence. The third is that there is no third status. Let p stand for “L is true”. Premises one and two give that p holds exactly when L is not true, that is, when ¬p. So the conclusion is the biconditional below.
The lab builds its table. There are two rows, p true and p false, and in each the biconditional is false. A sentence false in every row is a contradiction, and this one is the form from the barber, so the premises of the liar argument cannot all be true.
The two halves can be read separately. The formula p → ¬p says that if L is true then it is not, and it is true in exactly one row, the row where p is false. The formula ¬p → p says that if L is not true then it is, and it is true in exactly one row, the one where p is true. Each half is satisfiable alone, and they are satisfiable in different rows, which is why the pair is not. The formula p ∨ ¬p is true in both rows, and it is the form of the third premise: the table gives it no trouble, and the trouble is that it cannot be kept with the other two.
This is why the common first reply fails. Someone says the liar is simply false. That is the row where p is false. In that row L is not true, which is exactly what L says, so things are as L says and by the schema L is true. The reply starts in a row and ends in the other. Saying it is simply true fails the same way in reverse. The table has no row left to put the liar in.
Saying the sentence is neither true nor false does not escape either, and the wording of L shows why. It says it is not true. If it is neither, then it is not true, which is what it says, so things are as it says and it is true after all. A gap between true and false is stepped around by “not true”, unless the third premise is the one given up and the schema is restricted to match.
The exit that defines the field is Tarski's. A language cannot contain its own truth predicate. The word “true” as applied to sentences of one language belongs to a richer language, which can talk about the first; to speak of the truth of sentences of that richer language one needs a richer one still. The first premise is where this bites: no sentence of the lower language can say of itself that it is not true, because the lower language has no such word. The liar is not false or gappy. It is not a sentence of any language in the hierarchy, and the table is silent about sentences that are never formed.
The price is real and should be stated at its strongest. English has one word “true”, and it is used without a subscript about sentences that contain it. The hierarchy is a rule that English does not follow, and its defenders say that the rule is a repair and not a description. Two other exits are on offer. One gives up the third premise and makes room for sentences that are neither, at the cost of the strengthened version above. The other accepts the conclusion, so that L is both true and not true; it needs a consequence relation in which a contradiction does not entail everything, and the lab does not implement one. Which of the three is the least costly is the question the lesson leaves open.
Interactive
Every assignment, evaluated
Method
Four moves. The first three are mechanical, and the fourth is where the philosophy is.
Write down the sentence L and let p stand for “L is true”. The sentence says that it is not true, which is ¬p.
A sentence is true exactly when things are as it says. For L that gives p ↔ ¬p.
Build the two rows. If the biconditional is false in both, the three premises cannot all hold.
Forbid the self-referring sentence, allow a status other than true or not true, or accept the contradiction. Whichever you pick, write what it costs and what the strengthened liar says to it.
Let p stand for “L is true”. Read the two rows of the biconditional the argument ends with.
The table settles that the three premises are jointly impossible. It does not settle which is false, and each reply to the liar is an answer to that second question.
Given the liar or a variant, you should be able to let a letter stand for its truth, write the biconditional the schema yields, say from a table whether any row satisfies it, and state which premise a given reply rejects and the sentence that tests that reply.
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