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Topos Uranos

Summary

This lesson delimits the object of mathematical logic: not the truth of particular statements, which belongs to each science, but the form of reasoning and validity, that is, the relation between premises and conclusion that guarantees that the truth of the former passes to the latter. It defines proposition, argument, premise, conclusion, form and counterexample; it distinguishes validity from truth by means of valid arguments with a false conclusion and invalid arguments with a true conclusion; it shows why natural language, ambiguous and able to speak about itself, does not suffice to study validity rigorously, and it establishes the distinction between object language and metalanguage, which governs the whole course. It then sets out, with their explicit hypotheses and their proofs, the paradoxes of the liar, of Berry and of the barber, together with Russell's, which the barber anticipates, and explains how a formal language avoids them or turns them into theorems. It ends with the four pillars of mathematical logic and with the map of the course.

Learning objectives

  1. Distinguish the validity of an argument from the truth of its premises and of its conclusion, and refute an invalid form by constructing a counterexample.
  2. Recognize ambiguities of natural language and make each reading of an ambiguous statement precise.
  3. Distinguish the object language from the metalanguage, and the use of an expression from its mention.
  4. Analyze a paradox: isolate its hypotheses, deduce the contradiction and decide which hypothesis must be abandoned.

What logic studies

All knowledge that is not obtained by direct observation is obtained by reasoning: from certain assertions that are accepted one passes to another that follows from them. In geometry, whoever knows that the angles of a triangle add up to two right angles concludes that no triangle has two obtuse angles; in medicine, whoever knows that a certain disease produces fever, and that the person examined has none, rules that disease out. Logic is the discipline that studies when that step is legitimate, and mathematical logic is the one that studies it with the methods of mathematics: defining its objects exactly, which are languages, formulas and deductions, and proving theorems about them.

To state precisely what it studies, it is convenient first to fix the vocabulary. A proposition is a declarative sentence of which it makes sense to say that it is true or that it is false: “77 is a prime number” is a true proposition; “every even number is greater than 55”, a false proposition. On the other hand, questions (“what time is it?”), commands (“close the door”) and expressions with free variables, such as “xx is even”, whose value depends on what xx designates, are not propositions.

An argument is a list of propositions, the premises, together with another proposition, the conclusion, which is presented as a consequence of them. The premises may be any propositions whatsoever: true, false, known or mere suppositions; the only thing that distinguishes them is the place they occupy in the argument. We shall write arguments with the premises on successive lines and the conclusion preceded by “therefore”:

ExampleAn argument

If a number is a multiple of 44, it is even. The number 2828 is a multiple of 44. Therefore, 2828 is even.

Its premises are “if a number is a multiple of 44, it is even” and “2828 is a multiple of 44”; its conclusion, “2828 is even”.

Logic is not a theory of truth

If this is so, logic does not have the task of finding out whether the premises of an argument are true. Whether 2828 is a multiple of 44 is decided by arithmetic; whether a person has a fever, by the thermometer; whether a battle took place in a certain year, by history. Nor does it fall to logic to define what truth is, a question that belongs to philosophy. Logic grants that propositions have a truth value, whatever it may be, and studies how those values are related to one another: in particular, when the truth of some propositions forces the truth of another. In other words, logic is not a theory of truth, but a theory of the transmission of truth.

Now, what transmits truth does not depend on the subject matter of the argument, but on its structure. Compare the previous argument with this one: “if an animal is a mammal, it breathes air; the whale is a mammal; therefore, the whale breathes air”. Both have the same form: “if something is AA, it is BB; cc is AA; therefore, cc is BB”. The form of an argument is what remains when its non-logical terms (numbers, animals, properties) are replaced by letters, keeping the logical words: “not”, “and”, “or”, “if… then”, “all”, “some”. Each argument obtained from a form by giving content to its letters is an instance of it.

A form is valid when none of its instances has all its premises true and its conclusion false; an argument is valid when its form is. An instance with true premises and a false conclusion is called a counterexample to the form: whence a form is invalid exactly when it has a counterexample.

In the proofs of this lesson we shall use a shorthand of English: V(A)V(A) abbreviates “AA is true” and F(A)F(A), “AA is false”; ¬\neg abbreviates “not”, ∧\wedge abbreviates “and”, ∨\vee abbreviates “or”, ⇒\Rightarrow abbreviates “if… then” and ⇔\Leftrightarrow abbreviates “if and only if”; ∀\forall is read “for all” and ∃\exists, “there exists”. These signs are here mere abbreviations of words of the metalanguage, and the double arrows are reserved throughout the course for that use. In the next lesson the single-stroke signs, →\rightarrow and ↔\leftrightarrow, will appear, together with ¬\neg, ∧\wedge and ∨\vee, as abbreviations of a formal language, with a meaning that the course will fix exactly; the difference in stroke between the arrows will recall at which level one is speaking.

PropositionWhat validity guarantees

If an argument is valid and all its premises are true, its conclusion is true. Consequently, if a valid argument has a false conclusion, some one of its premises is false.

Proof

  1. V(A1)∧⋯∧V(An)\dato{p}{V(A_1) \wedge \cdots \wedge V(A_n)}

    Let there be a valid argument with premises A1,…,AnA_1, \ldots, A_n and conclusion CC, and suppose that all the premises are true.

  2. A1,…,An,C\resaltar{A_1, \ldots, A_n, C}

    Every argument is an instance of its own form: it suffices to give the letters of the form the content they had in the argument.

  3. ¬(V(A1)∧⋯∧V(An)∧¬V(C))\dato{v}{\neg\left( V(A_1) \wedge \cdots \wedge V(A_n) \wedge \resaltar{\neg V(C)} \right)}

    By the definition of validity, no instance of that form has its premises true and its conclusion false; in particular, the argument itself does not.

  4. V(C)\resaltar{V(C)}

    Since the premises are true, the only thing that can fail in the negated conjunction is the falsity of the conclusion: the conclusion is true.

  5. ¬V(C)⇒¬(V(A1)∧⋯∧V(An))\neg V(C) \Rightarrow \neg\left( V(A_1) \wedge \cdots \wedge V(A_n) \right)

    If, on the contrary, the conclusion is false, the premises cannot all be true, because then the conclusion would be.

The proposition states what validity guarantees; it is worth seeing just as clearly what it does not guarantee. Of the eight possible combinations of the validity of the argument, the truth of all its premises and the truth of its conclusion, only one is excluded, that of the valid argument with true premises and a false conclusion; the others all occur.

ExampleValid, with a false conclusion

Every prime number is odd. The number 22 is prime. Therefore, 22 is odd.

The form is the same as that of the first argument of the lesson, so the argument is valid; nevertheless, its conclusion is false. The preceding proposition points out where the defect lies: some premise is false, and indeed the first one is, because 22 is prime and even. A valid argument does not produce truths; it only preserves those it receives.

ExampleInvalid, with a true conclusion

If a number is a multiple of 44, it is even. The number 88 is even. Therefore, 88 is a multiple of 44.

Premises and conclusion are true, and even so the argument is invalid. Its form, “if something is AA, it is BB; cc is BB; therefore, cc is AA”, has a counterexample: with the number 66 in place of 88, the premises remain true (“if a number is a multiple of 44, it is even”; “66 is even”) and the conclusion, “66 is a multiple of 44”, is false. The truth of the original conclusion does not come from the premises, but from arithmetic: 8=4⋅28 = 4 \cdot 2. This invalid form, frequent in everyday reasoning, is called affirming the consequent.

From this follows the practical rule that the lesson will exercise: to show that an argument is invalid it is not enough to point out that some premise or the conclusion is false, and to show that it is valid it is not enough that everything in it be true; the former requires exhibiting a counterexample to its form, and the latter, a reasoning that holds for all its instances.

Natural language and formal language

The definition of validity rests on two operations: recognizing the form of an argument and deciding whether a proposition is true or false. Both presuppose that each sentence has a single reading. Natural language, English, Chinese or Arabic, which human beings learn without formal instruction and use to communicate, does not meet that condition: its flexibility, which makes it fit for poetry, conversation and nuance, also makes it ambiguous.

  • Lexical ambiguity: a word has several meanings. “I sat down by the bank” may refer to the edge of a river or to a financial institution.
  • Syntactic ambiguity: a sentence admits several structures. “Twice three plus four” may be 2⋅3+4=102 \cdot 3 + 4 = 10 or 2⋅(3+4)=142 \cdot (3 + 4) = 14.
  • Scope ambiguity: it is unclear which part of the sentence a logical word affects. “All the students did not pass” may mean that none passed or that not all passed.
  • Imprecise connectives: “or” sometimes allows both alternatives to hold (“a degree or experience is required”) and sometimes excludes it (“coffee or tea”); “if… then” sometimes expresses a cause, sometimes a rule, sometimes a mere succession.

To ambiguity is added a deeper feature: natural language can speak about itself. It contains the words “sentence”, “true” and “define”, and with them it can form sentences that refer to themselves, such as “this sentence has five words”. This self-reference is, in general, harmless; but, as will be seen, combined with the notion of truth it produces sentences that can be neither true nor false.

A formal language is designed to eliminate both sources of difficulty. One fixes an alphabet, a collection of symbols, and some formation rules that say, leaving no room for interpretation, which sequences of symbols are well-formed expressions; the meaning of each expression is fixed afterwards by equally precise definitions. Thus, asking whether a sequence of symbols belongs to the language is a question that is settled mechanically, by checking rules, and each expression is read in only one way. The language of this course, which the lesson on the language of propositional logic builds, has an alphabet of only two symbols, the circle ∘\sigO and the disk ∙\sigI, and a single formation rule.

The price is flexibility: a formal language is no good for conversation, and each formal language expresses only what its rules allow. It is not true, however, that all expressive power is sacrificed: the formal languages of first-order logic suffice to write almost all of mathematics. What is sacrificed is ambiguity and uncontrolled self-reference; what is gained is the possibility of proving theorems about the language itself.

Object language and metalanguage

To study a language one has to speak about it, and to speak about it another language is needed. The object language is the language under study; the metalanguage is the language in which it is studied. In a Latin grammar written in English, Latin is the object language and English the metalanguage. In this course, the object language will be the formal language of circles and disks, and the metalanguage will be English, enriched with mathematical signs and with Greek letters, φ\varphi, ψ\psi, which will designate arbitrary formulas of the object language without belonging to it.

The distinction is central for a precise reason: the assertions that the course will prove about the formal language (that every formula is read in only one way, that everything derivable is true) are not formulas of the formal language, but sentences of the metalanguage. That is why they will be called metatheorems, to distinguish them from the theorems that the deductive system produces within the object language.

An elementary form of the distinction is the one that separates the use of an expression from its mention. In “Rome is a city” the word is used to speak about the city; in “‘Rome’ has four letters” the word is mentioned, and the quotation marks indicate it: the sentence speaks about the word, not about the city. Confusing the two produces absurd statements, such as “Rome has four letters”, or fallacious arguments, such as “Rome is a city; ‘Rome’ is a four-letter word; therefore, a city is a four-letter word”.

ExampleThree levels

Let us take Latin as the object language. The sentence “Roma aeterna est” belongs to the object language. The sentence “‘Roma aeterna est’ has three words” belongs to the metalanguage, English, and is true. The sentence “the preceding sentence is written in English” speaks about a sentence of the metalanguage: it belongs to a third level, the metalanguage of the metalanguage, even though it is written in the same language. Nothing prevents one and the same language from serving at several levels; what matters is not to lose sight of what is being spoken about.

The paradoxes of language

A contradiction is an assertion together with its negation, such as V(A)∧¬V(A)V(A) \wedge \neg V(A). A paradox is a piece of reasoning that, starting from hypotheses that seem acceptable and advancing by steps that seem valid, arrives at a contradiction. By the proposition on what validity guarantees, if the steps are really valid, some hypothesis is false: resolving a paradox consists in finding which. One should not confuse paradox with ambiguity, which is merely the multiplicity of readings, nor with what is merely surprising.

The liar

The oldest paradox of this kind is attributed to Eubulides of Miletus, in the fourth century before the common era. In its precise form it does not speak of lying, which presupposes the intention to deceive, but of falsity: let mm be the sentence “the sentence mm is false”. To obtain the contradiction two hypotheses suffice, which it is worth writing out explicitly.

  • Bivalence (B): every proposition is true or false, and not both; being false is equivalent to not being true.
  • Truth schema (T): a sentence is true if and only if what it says is the case. For example, “snow is white” is true if and only if snow is white.
TheoremThe liar paradox

The hypotheses of bivalence and of the truth schema, applied to the sentence mm, lead to a contradiction: there is no way of assigning to mm a truth value that respects them. In the figure, the sentence mm points to itself, and the arrows run through the two possibilities.

Proof

  1. F(m)⇔¬V(m)\dato{B}{F(m) \Leftrightarrow \neg V(m)}

    By bivalence, mm is true or false, and not both: being false is not being true.

  2. V(m)⇔F(m)\dato{T}{V(m) \Leftrightarrow \resaltar{F(m)}}

    What mm says is that mm is false. By the truth schema, mm is true exactly when what it says is the case.

  3. V(m)⇔¬V(m)V(m) \Leftrightarrow \resaltar{\neg V(m)}

    Chaining both equivalences, the truth of mm is equivalent to its own negation.

  4. V(m)⇒¬V(m),¬V(m)V(m) \Rightarrow \neg V(m), \quad \resaltar{\neg V(m)}

    If mm were true, by the equivalence read from left to right it would not be: the supposition refutes itself, and mm is not true.

  5. V(m)∧¬V(m)\resaltar{V(m) \wedge \neg V(m)}

    But then, by the equivalence read from right to left, mm is true: one has at once an assertion and its negation.

The theorem does not say that logic is contradictory, but that the two hypotheses cannot be maintained together for a language that contains the sentence mm. There are, therefore, three ways out: abandoning bivalence and admitting sentences without a truth value; restricting the truth schema; or preventing the language from forming the sentence mm. The third is that of Alfred Tarski, in 1933: the predicate “is true” of a language does not belong to that language, but to its metalanguage. In such a language, mm is not a well-formed sentence, and the paradox cannot be stated.

Berry's paradox

Published by Bertrand Russell in 1906, who attributed it to George Berry, a librarian at Oxford, this paradox does not use the word “true”, but the word “define”. An English phrase is said to define a natural number when it describes it in such a way that the number can be identified without ambiguity: “the smallest prime number greater than ten” defines 1111. Let us consider the phrase “the smallest natural number that cannot be defined in fewer than twenty words”, which must be read just as it is written, and let us call bb the number it purports to define. The natural numbers, in this course, are 1,2,3,…1, 2, 3, \ldots

TheoremBerry's paradox

If the vocabulary of English is finite and the relation “the phrase defines the number” is well determined, a contradiction is reached: the number bb exists, and at the same time it can and cannot be defined in fewer than twenty words. In the figure, the short phrases are sent to the numbers they define, which are few; the first number left free is bb.

Proof

  1. N≤k+k2+⋯+k19\dato{N}{N \leq k + k^2 + \cdots + k^{19}}

    Let kk be the number of words in the dictionary. The phrases of one word are at most kk; those of two, at most k2k^2; and so on up to those of nineteen. The phrases of fewer than twenty words are, therefore, a finite number NN.

  2. D={d1,…,dr},r≤N\dato{D}{D = \{ d_1, \ldots, d_r \}, \quad r \leq N}

    Each phrase defines at most one number. Let DD be the set of the naturals that some phrase of fewer than twenty words defines: it has at most NN elements.

  3. (∃n≤N+1) (n∉D),b∉D(\exists n \leq N + 1)\,(n \notin D), \quad \resaltar{b \notin D}

    Among the N+1N + 1 numbers 1,2,…,N+11, 2, \ldots, N + 1 there is some one that is not in DD. Running through them in order, the first one that is not in DD is the smallest natural number that cannot be defined in fewer than twenty words: all the preceding ones are in DD.

  4. 13<20,b∈D13 < 20, \quad \resaltar{b \in D}

    But the phrase “the smallest natural number that cannot be defined in fewer than twenty words” has thirteen words, and defines precisely bb.

  5. b∈D∧b∉D\resaltar{b \in D \wedge b \notin D}

    One has at once an assertion and its negation.

The usual exposition, which confines itself to observing that the phrase has fewer than twenty words, omits half of the argument: that the number bb exists, which requires the finiteness of the vocabulary and the choice of the first number missing. The hypothesis that fails is the second: “define” is not a precise relation of English, because the phrase that defines bb uses the very word “define”, and its sense depends on which phrases count as definitions. In a formal language LL one can make precise what “definable in LL” means, but that notion is expressible only in the metalanguage; the phrase that defines bb does not belong to LL, and the contradiction disappears: bb is not definable in LL with few symbols, although it is so in the metalanguage.

The barber and Russell's set

The barber paradox is usually told thus: in a village there is a barber, a man of the village, who shaves all the men of the village who do not shave themselves, and only them. Does the barber shave himself? The words “and only them” are essential: a barber who shaved all those who do not shave themselves, without the restriction, could also shave himself without any contradiction. The paradox is a particular case of a general lemma, of diagonal form, which it is worth proving once and for all.

LemmaThe diagonal lemma

Let DD be a collection of objects and RR a relation between them, such that, for each pair xx, yy of objects of DD, the assertion R(x,y)R(x, y) is true or false. Then there is no object bb of DD such that, for every xx of DD, R(b,x)R(b, x) holds if and only if R(x,x)R(x, x) does not hold. In the figure, each row is an object xx and each column an object yy, and a cell is filled when R(x,y)R(x, y) holds; the row we are looking for for bb would have to invert the diagonal, and that is why it differs from each row in that row's diagonal cell.

Proof

  1. (∀x∈D) (R(b,x)⇔¬R(x,x))\dato{b}{(\forall x \in D)\,\left( R(b, x) \Leftrightarrow \neg R(x, x) \right)}

    Suppose there exists such a bb in DD.

  2. R(b,b)⇔¬R(b,b)R(b, \resaltar{b}) \Leftrightarrow \neg R(\resaltar{b}, \resaltar{b})

    The condition holds for every xx of DD, and bb is one of them: we can take x=bx = b.

  3. R(b,b)⇒¬R(b,b)R(b, b) \Rightarrow \resaltar{\neg R(b, b)}

    If R(b,b)R(b, b) holds, by the equivalence it does not hold: a contradiction.

  4. ¬R(b,b)⇒R(b,b)\neg R(b, b) \Rightarrow \resaltar{R(b, b)}

    If R(b,b)R(b, b) does not hold, by the equivalence it holds: a contradiction as well.

  5. ¬(∃b∈D) (∀x∈D) (R(b,x)⇔¬R(x,x))\neg (\exists b \in D)\,(\forall x \in D)\,\left( R(b, x) \Leftrightarrow \neg R(x, x) \right)

    Since one of the two cases must occur, the supposition is untenable: there is no such bb.

CorollaryThere is no such barber

In no village is there a man of the village who shaves all the men of the village who do not shave themselves, and only them.

Proof

  1. (∀x∈D) (R(b,x)⇔¬R(x,x))(\forall x \in D)\,\left( R(b, x) \Leftrightarrow \neg R(x, x) \right)

    Let DD be the collection of the men of the village and R(x,y)R(x, y) the relation “xx shaves yy”. The barber bb is a man of the village, so he is in DD; that he is a barber as described means that he shaves each man of the village exactly when that man does not shave himself.

  2. ¬(∃b∈D) (∀x∈D) (R(b,x)⇔¬R(x,x))\resaltar{\neg (\exists b \in D)}\,(\forall x \in D)\,\left( R(b, x) \Leftrightarrow \neg R(x, x) \right)

    By the diagonal lemma, no man of the village satisfies that condition: the barber described does not exist. The story is not a paradox, but the proof that its protagonist is impossible.

The conclusion about the barber is harmless: nobody had claimed that such a barber existed. The same does not happen with its version in set theory. In 1902, Russell wrote to Gottlob Frege that his system, which aimed to found arithmetic on logic, admitted the following comprehension principle: for every property there is a set whose elements are exactly the objects that have it. Applied to the property of not belonging to itself, the principle asserts that there exists the set

Z={x:x∉x}Z = \{ x : x \notin x \}
CorollaryRussell's paradox

The comprehension principle, without restrictions, is contradictory.

Proof

  1. (∀x∈D) (x∈Z⇔x∉x)\dato{Z}{(\forall x \in D)\,\left( x \in Z \Leftrightarrow x \notin x \right)}

    Let DD be the collection of all sets and R(x,y)R(x, y) the relation “yy belongs to xx”. The comprehension principle, applied to the property x∉xx \notin x, asserts that there exists a set ZZ whose elements are the sets that do not belong to themselves.

  2. (∀x∈D) (R(Z,x)⇔¬R(x,x))(\forall x \in D)\,\left( \resaltar{R(Z, x)} \Leftrightarrow \resaltar{\neg R(x, x)} \right)

    With the relation chosen, the condition on ZZ is exactly that of the diagonal lemma, with ZZ in place of bb: x∈Zx \in Z is R(Z,x)R(Z, x), and x∉xx \notin x is ¬R(x,x)\neg R(x, x).

  3. Z∈Z⇔Z∉Z\resaltar{Z \in Z \Leftrightarrow Z \notin Z}

    The set ZZ is a set, so it is in DD; by the diagonal lemma, no such ZZ exists, and taking x=Zx = Z one obtains the impossible equivalence. But the principle asserts that it exists. The principle therefore implies a contradiction.

Unlike the barber, here the existence of ZZ was not a supposition in a story, but a principle of the theory. Russell's paradox is not resolved by declaring that ZZ does not exist, but by reforming the theory: Zermelo's axiomatic set theory, of 1908, only allows forming, with a property, subsets of an already given set, and with that restriction the contradiction is not deduced.

What is not a paradox

The observation that the empty set exists although it has no elements, as if an existing object had to be made of existing objects, has sometimes been called the “paradox of the empty set”. There is no contradiction in it whatsoever: a set is not made of its elements as a wall is made of its bricks, but is determined by them, and the collection of the elements of a set may have none, just as an empty box is still a box. That a fact is surprising does not make it paradoxical; a paradox requires deducing a contradiction, and here none is deduced.

How a formal language deals with them

From the foregoing it follows that a formal language does not avoid the paradoxes by magic, but in two precise ways. The first is that it may lack the resources that produce them. The language of propositional logic, which this course builds, has no sentences that refer to themselves and no truth predicate: in it the liar sentence cannot be written, and its semantics assigns to each formula, under each assignment of values to the variables, a single truth value. The second is that, in richer languages, the paradox becomes a theorem: the liar's reasoning, carried out rigorously, proves that the truth of a sufficiently expressive language is not definable in that same language (Tarski); that of the barber and of Russell proves that certain collections are not sets; and arguments related to Berry's lie at the heart of Kurt Gödel's incompleteness theorems, of 1931. Thus, what in natural language is a perplexity, in a formal language is information.

The four pillars of mathematical logic

Mathematical logic was born at the end of the nineteenth century and the beginning of the twentieth, with Frege, Peano, Russell and David Hilbert, as the answer to a precise question: on what foundations mathematics rests and whether those foundations are free from contradiction. Today it is divided into four great branches, which constantly intertwine.

  • Proof theory: it studies formal proofs as mathematical objects, that is, as finite sequences of formulas that follow explicit rules; it asks what can be proved in a system, whether a system is consistent and how some proofs are transformed into others. The deduction unit of this course is a first contact with it.
  • Model theory: it studies the relation between a formal language and the structures that interpret it, its models; it asks which structures satisfy a set of formulas and which properties of structures can be expressed in a given language. The semantics unit, with its assignments of truth values, is its most elementary version.
  • Set theory: it studies sets with precise axioms, such as those of Zermelo and Fraenkel, in which almost all of mathematics can be written; it was born, in good part, from the need to avoid Russell's paradox.
  • Computability theory: it studies which problems can be solved by a mechanical procedure and which cannot. Propositional logic is decidable, as the course will show: there is a procedure that decides in a finite number of steps whether a formula is valid; predicate logic, on the other hand, is not.

No pillar is studied in depth without touching the others: completeness, a central result of this course, is at once a theorem of proof theory and of model theory, and the decidability obtained from it belongs to computability.

The map of the course

The course travels, for the simplest language of logic, the propositional one, the complete path of a logical theory, in four units.

  1. Language: syntax. The object language is built with its two-symbol alphabet and its formation rule, it is proved that each formula is read in only one way, one learns to prove by induction, and statements and arguments of natural language are translated into the formal language, pointing out what the translation preserves and what it loses.
  2. Deduction: a system of axioms and rules of inference is fixed, and it is defined what it means for a formula φ\varphi to be deduced from a set of formulas Γ\Gamma, which is written Γ⊢φ\Gamma \vdash \varphi. It is the syntactic face of validity: manipulating symbols according to rules, without attending to their meaning.
  3. Semantics: each formula is given a truth value under each assignment, and it is defined what it means for φ\varphi to be a consequence of Γ\Gamma, which is written Γ⊨φ\Gamma \models \varphi: every assignment that makes the formulas of Γ\Gamma true makes φ\varphi true. It is the semantic face of validity, the definition of this lesson made precise.
  4. Metatheory: it is proved that both faces coincide. Soundness says that everything derivable is a consequence, and completeness, that every consequence is derivable. Afterwards compactness is studied, together with a mechanical method, resolution, for deciding whether a set of formulas can be satisfied.

If this is so, the course is the first complete example of a mathematical theory whose object is reasoning itself; and all its assertions about the formal language will be, as has been seen in this lesson, sentences of the metalanguage proved with the rigor of mathematics.

Solved problems

The following problems exercise, in increasing difficulty, the distinctions of the lesson: refuting an argument whose sentences are all true, separating the readings of an ambiguous statement, classifying expressions of the object language and of the metalanguage, and analyzing two pieces of reasoning of paradoxical appearance to decide whether they are paradoxes.

Solved problem 1All true, and still invalid

Show that the following argument is invalid, even though its premises and its conclusion are true: “every multiple of 66 is even; the number 1212 is even; therefore, 1212 is a multiple of 66”.

Solution

  1. (∀x) (S(x)⇒P(x)),P(12),S(12)(\forall x)\,(S(x) \Rightarrow P(x)), \quad P(12), \quad S(12)

    We abbreviate “xx is a multiple of 66” by S(x)S(x) and “xx is even” by P(x)P(x). The premises and the conclusion are true, because 12=6⋅212 = 6 \cdot 2.

  2. (∀x) (S(x)⇒P(x)),P(c),S(c)(\forall x)\,(S(x) \Rightarrow P(x)), \quad P(\resaltar{c}), \quad S(\resaltar{c})

    The form is obtained by replacing the name 1212 with a letter cc: two premises and a conclusion.

  3. (∀x) (S(x)⇒P(x)),P(4)(\forall x)\,(S(x) \Rightarrow P(x)), \quad P(\resaltar{4})

    We look for an instance with true premises: the first does not depend on cc, and the second asks for an even number. We take c=4c = 4.

  4. ¬S(4)\neg S(\resaltar{4})

    The conclusion of this instance is false, because 44 is not a multiple of 66.

  5. 12=6⋅2\resaltar{12 = 6 \cdot 2}

    The form has a counterexample: it is invalid, and so is the argument. The truth of its conclusion comes from arithmetic, not from the premises.

Solved problem 2A scope ambiguity

Let LL be the list of numbers 44, 66, 77. Show that the sentence “all the numbers of LL are not prime” has two readings, and that with this list one is true and the other false.

Solution

  1. ¬Q(4)∧¬Q(6)∧¬Q(7)\dato{a}{\neg Q(4) \wedge \neg Q(6) \wedge \neg Q(7)}

    We abbreviate “xx is prime” by Q(x)Q(x). First reading: the negation affects each number, and the sentence says that none is prime.

  2. ¬(Q(4)∧Q(6)∧Q(7))\dato{n}{\neg\left( Q(4) \wedge Q(6) \wedge Q(7) \right)}

    Second reading: the negation affects the whole assertion, and the sentence says that not all are prime.

  3. Q(7),¬(¬Q(4)∧¬Q(6)∧¬Q(7))Q(7), \quad \neg\left( \neg Q(4) \wedge \neg Q(6) \wedge \resaltar{\neg Q(7)} \right)

    Since 77 is prime, the last term of the first reading is false, and with it the whole conjunction: the first reading is false.

  4. ¬Q(4),¬(Q(4)∧Q(6)∧Q(7))\neg Q(4), \quad \neg\left( \resaltar{Q(4)} \wedge Q(6) \wedge Q(7) \right)

    Since 44 is not prime, the conjunction of the second reading is false, and its negation true: the second reading is true.

  5. ¬Q(4)∧¬Q(6)∧¬Q(7)⇒¬(Q(4)∧Q(6)∧Q(7))\resaltar{\neg Q(4) \wedge \neg Q(6) \wedge \neg Q(7)} \Rightarrow \neg\left( Q(4) \wedge Q(6) \wedge Q(7) \right)

    The two readings are not equivalent: the first implies the second, but not conversely, and this list shows it. In mathematics, the second is written “not all the numbers of LL are prime”, and the first, “no number of LL is prime”.

Solved problem 3Object language and metalanguage

Let us take as object language the sequences of the symbols ∘\sigO and ∙\sigI, and as metalanguage English, with Greek letters that designate arbitrary sequences. Classify each of the following expressions and decide, where appropriate, whether it is true: (a) ∘∘∙\sigO\sigO\sigI; (b) “∘∘∙\sigO\sigO\sigI has three symbols”; (c) “for every sequence φ\varphi, the sequence ∙φ\sigI\varphi begins with a disk”; (d) “∘∘∙\sigO\sigO\sigI is true”.

Solution

  1. ∘∘∙\resaltar{\sigO\sigO\sigI}

    Expression (a) belongs to the object language: it is a sequence of symbols, not an English sentence, and it is neither true nor false; it is spoken about, but it asserts nothing.

  2. ∘  ∘  ∙(3)\sigO \; \sigO \; \sigI \quad (3)

    Expression (b) is a sentence of the metalanguage that mentions sequence (a) and counts its symbols: it is true.

  3. ∙φ\resaltar{\sigI}\varphi

    Expression (c) is a sentence of the metalanguage. The letter φ\varphi is not a symbol of the object language, but a variable of the metalanguage that ranges over its sequences; the sentence is true, because ∙φ\sigI\varphi begins, by construction, with the disk.

  4. v2=∘∘∙v_2 = \sigO\sigO\sigI

    Expression (d) belongs to the metalanguage, but it does not yet have a truth value: the sequence ∘∘∙\sigO\sigO\sigI will be a variable of the language of the course, and a variable is neither true nor false by itself, but only when an assignment, in the semantics unit, gives it a value. In the next lesson it will be called v2v_2.

Solved problem 4Epimenides is not the liar

On a blackboard there is written the sentence ee, perhaps among others: “every sentence written on this blackboard is false”. Granting bivalence and the truth schema, prove that ee is false and that there is some other true sentence on the blackboard; and deduce that there is a paradox exactly when no other sentence on the blackboard is true, as happens if ee is the only one.

Solution

  1. V(e)⇔(∀x∈O) F(x)\dato{T}{V(e) \Leftrightarrow (\forall x \in O)\,F(x)}

    Let OO be the set of the sentences on the blackboard; ee is one of them. By the truth schema, ee is true if and only if all the sentences of OO are false.

  2. V(e)⇒F(e)⇒¬V(e)V(e) \Rightarrow \resaltar{F(e)} \Rightarrow \neg V(e)

    If ee were true, all the sentences of OO would be false, and ee is in OO: ee would be false, and by bivalence it would not be true. The supposition refutes itself.

  3. F(e)\resaltar{F(e)}

    Therefore, ee is not true; by bivalence, it is false.

  4. (∃x∈O) (x≠e∧V(x))(\exists x \in O)\,\left( x \neq e \wedge V(x) \right)

    By the truth schema, that ee is false means that not all the sentences of OO are false: some one is true, and it is not ee, which is false.

  5. ¬(∃x∈O) (x≠e∧V(x))⇒V(e)∧¬V(e)\neg (\exists x \in O)\,\left( x \neq e \wedge V(x) \right) \Rightarrow \resaltar{V(e) \wedge \neg V(e)}

    If there is another true sentence on the blackboard, everything is coherent: ee is simply false. If, on the other hand, no other sentence is true, there is no xx different from ee in OO that is true, and a contradiction is obtained, as in the liar; this happens, in particular, when ee is the only sentence on the blackboard. Whether the paradox appears depends, therefore, on an empirical fact: what is written on the blackboard.

Solved problem 5The heterological

An adjective is autological if it applies to itself (“polysyllabic” is polysyllabic; “short” is a short word) and heterological if it does not (“monosyllabic” is not monosyllabic; “long” is not a long word). Prove that, if “heterological” is an adjective and the relation “applies to” is well determined for all adjectives, a contradiction is reached; and decide which hypothesis to abandon.

Solution

  1. R(x,x)\dato{R}{R(x, x)}

    Let DD be the collection of the adjectives of English and R(x,y)R(x, y) the relation “the adjective xx applies to the word yy”. An adjective xx is autological if R(x,x)R(x, x) and heterological if ¬R(x,x)\neg R(x, x).

  2. (∀x∈D) (R(h,x)⇔¬R(x,x))\dato{h}{(\forall x \in D)\,\left( R(h, x) \Leftrightarrow \neg R(x, x) \right)}

    Let hh be the adjective “heterological”. By its definition, it applies to an adjective exactly when that adjective does not apply to itself.

  3. R(h,h)⇔¬R(h,h)R(\resaltar{h}, \resaltar{h}) \Leftrightarrow \neg R(\resaltar{h}, \resaltar{h})

    If hh is an adjective, it is in DD, and its condition is that of the diagonal lemma. Taking x=hx = h, the impossible equivalence is obtained.

  4. h∈D⇒h∉Dh \in D \Rightarrow \resaltar{h \notin D}

    By the diagonal lemma, no object of DD satisfies the condition of hh. If hh is in DD, there is a contradiction.

  5. h∉D\resaltar{h \notin D}

    The hypothesis that fails is the second: “applies to” is not a precise relation for adjectives that, like “heterological”, are themselves defined by that same relation. In a formal language, the relation RR would be defined in the metalanguage, and the adjective hh would not belong to the object language, so that the condition h∈Dh \in D would be false and there would be no contradiction.

Proposed problems

The first two problems exercise validity and ambiguity; the third, use and mention; the last two ask you to analyze a new paradox and to apply the diagonal lemma outside linguistics. Each one has a hint, which is best opened only after having attempted the problem.

Proposed problem 1Two similar forms

Consider the following two forms: (a) “if pp, then qq; not qq; therefore, not pp”; (b) “if pp, then qq; not pp; therefore, not qq”. Decide which is valid and which is not; for the invalid one, construct a counterexample with statements about numbers, and for the valid one, explain why no instance can be a counterexample.

Hint

For form (b), look for a true implication between properties of numbers, such as “if nn is a multiple of 44, it is even”, and a number that does not have the first property but does have the second. For form (a), suppose that the premises are true and ask yourself what would happen if pp were true.

Answer

Form (a), called modus tollens, is valid: if the premises are true and pp were true, qq would be true as well, against the second premise. Form (b), denying the antecedent, is invalid: “if 66 is a multiple of 44, 66 is even; 66 is not a multiple of 44; therefore, 66 is not even” has true premises and a false conclusion.

Proposed problem 2Passport, identity card and visa

A notice says: “to enter, a passport or a national identity card and a visa are required”. Write its two readings with the letters pp (passport), dd (national identity card) and vv (visa), and describe a traveler whom one reading lets in and the other does not.

Hint

The question is whether the visa goes only with the national identity card or with either of the two documents. Try a traveler who has a single document.

Answer

The readings are p∨(d∧v)p \vee (d \wedge v) and (p∨d)∧v(p \vee d) \wedge v. A traveler with a passport and without a visa can enter according to the first and not according to the second.

Proposed problem 3Quotation marks

Decide whether each sentence speaks about the city or about the word, and whether it is true or false: (a) “Rome has four letters”; (b) “‘Rome’ has four letters”; (c) “‘Rome’ is the capital of Italy”; (d) “the sentence ‘Rome is the capital of Italy’ contains the word ‘Rome’”.

Hint

Quotation marks turn an expression into the name of that expression. Ask yourself, in each case, what object the subject of the sentence designates: a city, a word or a sentence.

Answer

(a) It speaks about the city and is false, because a city has no letters. (b) It speaks about the word and is true. (c) It speaks about the word and is false, because a word is not a capital. (d) It speaks about a sentence and about a word, and is true.

Proposed problem 4The two sides of a card

On one side of a card the sentence AA is written: “sentence BB is true”; on the other side, the sentence BB: “sentence AA is false”. Granting bivalence and the truth schema, prove that there is no way of assigning truth values to AA and to BB. What happens if BB says, instead, “sentence AA is true”?

Hint

Write the truth schema for each sentence, and chain the equivalences until you obtain one that relates V(A)V(A) to itself. In the variant, do the same and check which assignments satisfy it.

Answer

In the first version one obtains V(A)⇔¬V(A)V(A) \Leftrightarrow \neg V(A), as in the liar, without any sentence speaking about itself. In the variant, V(A)⇔V(B)V(A) \Leftrightarrow V(B) admits two assignments, both true or both false: there is no contradiction, but nothing decides between them.

Worked solution

Solution

  1. V(A)⇔V(B)\dato{A}{V(A) \Leftrightarrow V(B)}

    By the truth schema, AA is true exactly when BB is.

  2. V(B)⇔F(A)⇔¬V(A)\dato{B}{V(B) \Leftrightarrow F(A) \Leftrightarrow \neg V(A)}

    By the truth schema, BB is true exactly when AA is false; by bivalence, being false is not being true.

  3. V(A)⇔¬V(A)V(A) \Leftrightarrow \resaltar{\neg V(A)}

    Chaining both equivalences, the truth of AA is equivalent to its own negation, and one reasons as in the liar: if AA is true, it is not, and if it is not, it is.

  4. V(A)⇔V(B)⇔V(A)V(A) \Leftrightarrow V(B) \Leftrightarrow V(A)

    In the variant, the truth schema for BB gives V(B)⇔V(A)V(B) \Leftrightarrow V(A), which together with the first equivalence only requires that both have the same value: there are two coherent assignments and no contradiction.

Proposed problem 5Cantor's diagonal

Let s1,s2,s3,…s_1, s_2, s_3, \ldots be any list of infinite sequences of zeros and ones, and let us write sx(y)s_x(y) for the yy-th term of the sequence sxs_x. Prove, with the diagonal lemma, that the sequence dd defined by d(y)=1d(y) = 1 when sy(y)=0s_y(y) = 0, and d(y)=0d(y) = 0 when sy(y)=1s_y(y) = 1, does not appear in the list. Deduce that no numbered list contains all the sequences of zeros and ones.

Hint

Take as collection DD the natural numbers and as relation R(x,y)R(x, y) the assertion sx(y)=1s_x(y) = 1. What condition would a number bb such that sb=ds_b = d satisfy?

Answer

If d=sbd = s_b held, the number bb would satisfy R(b,y)⇔¬R(y,y)R(b, y) \Leftrightarrow \neg R(y, y) for every yy, which the diagonal lemma forbids. Therefore dd differs from each sbs_b at least in the bb-th term, and no list contains all the sequences.

Worked solution

Solution

  1. R(x,y)⇔sx(y)=1\dato{R}{R(x, y) \Leftrightarrow s_x(y) = 1}

    Let DD be the set of the natural numbers and R(x,y)R(x, y) the assertion that the yy-th term of sxs_x equals 11.

  2. d(y)=1⇔¬(sy(y)=1)\dato{d}{d(y) = 1 \Leftrightarrow \neg\left( s_y(y) = 1 \right)}

    By its definition, d(y)d(y) equals 11 exactly when sy(y)s_y(y) does not equal 11.

  3. (∀y∈D) (R(b,y)⇔¬R(y,y))(\forall y \in D)\,\left( R(\resaltar{b}, y) \Leftrightarrow \neg R(y, y) \right)

    Suppose that dd appears in the list in place bb, that is, d=sbd = s_b. Then, for every yy, R(b,y)R(b, y) is equivalent to d(y)=1d(y) = 1, which is equivalent to ¬R(y,y)\neg R(y, y).

  4. d(b)≠sb(b)d(b) \neq s_b(b)

    By the diagonal lemma, no bb satisfies that condition: the sequence dd does not appear in the list. Concretely, it differs from sbs_b in the bb-th term.

  5. (∀b∈N) (d≠sb)(\forall b \in \mathbb{N})\,(d \neq s_b)

    Since the list was arbitrary, no numbered list contains all the sequences of zeros and ones: every list lacks, at least, its diagonal sequence.