Quantifiers - University of Nebraska Omaha

14 ANDRZEJ ROSLANOWSKI 

3. Quantifiers 

The propositional calculus may to some extend formalize how we 

carry out arguments, but it is not strong enough to be actually useful. 

Only after allowing the use of variables and quantifiers in our syntax 

we will receive a formal structure that is rich and applicable. But this 

makes the topic harder and it is the treatment of quantifiers in mathematical 

statements that is the hardest part in mathematical proof 

writing. This is also what makes mathematical arguments di cult. 

3.1. Syntax: an informal introduction. We fix a list of allowed 

variables, say, x, x 1 ,x 2 ,..., y, y 1 ,y 2 ,... and z,z 1 ,z 2 ,.... We think 

about these variables as denoting some unspecified objects (compare 

with the phrase “a man entered the room”). 

All formulas we consider are related to some mathematical reality. 

It is convenient to think that the variables stand for objects from this 

reality, thattheycomefromafixeduniverse. It is just to make some 

expressions more practical: if we are talking about equations, then our 

variables should denote numbers. But if we are interested in insects, 

then the universe of our variables is the collection of all insects. This 

means that the scope of variables and/or quantifiers is often restricted 

to some fixed non-empty set called the “universe” or “domain.” 

We will formalize this notion later, but we can imagine what a formula 

in variables x 1 ,...,x n may mean. It is just some (mathematical) 

expression “talking” about unspecified objects x 1 ,...,x n .Examplesof 

such expressions are 

(⇤) 1 x 1 is bigger than x 2 +2 

(⇤) 2 x 1 is a prime number 

(⇤) 3 x 2 is a multiple of x 1 

(⇤) 4 x 1 + x 2 + x 3 2 x 1 

= x 4 

or looking at less mathematical context 

(⇤) 5 x loves y 

(⇤) 6 z is tall 

(⇤) 7 x is father of y. 

Each of the above examples comes with the domain of involved variables, 

but often we do not mention it explicitly. What are the domains 

of the variables in the above examples? 

If I ask you if the formulas listed above are true or false, you might 

be somewhat confused. Is x 1 bigger than x 2 +2? Depends on the 

particular choice of x 1 ,x 2 , but they are “free” in our sentence, they 

can be anything. You may pick x 1 =1andx 2 =2andclaimthatthe 

formula is false. But I will take x 1 =2013andx 2 =2andIwillinsist

MATH 2230 — FALL 2013 15 

that it is true. Thus, if a formula has a “free” instance of a variable, 

then it does not make much sense to ask if it is true 12 . 

Why do I add the adjective “free” when talking about a variable? 

Note that not every occurrence of a variable is the same as any other. 

Compare the following two formulas: 

(a) x is a prime number bigger than 1000, 

(b) there is a prime number x such that x>1000. 

Again, if I ask you if the first formula is true, than you will reply 

“depends on x”. But the same question on the second formula will 

be clearly answered with “sure, this formula is true”! So what is the 

di↵erence? Both formulas involve a variable x, but 

• the first formula “speaks” about x, while 

• the second formula “speaks” about 1000, the x here could be 

totally eliminated. (We could have just say “there is a prime 

number bigger than 1000.) 

Look carefully what happens here. We have formula 

x is a prime number bigger than 1000 

(i.e., formula in (a)) and then, essentially, we add in front of it the 

phrase “there is an x such that” getting 

there is an x such that x is a prime number bigger than 1000 

(and you should agree that this is the formula in (b)). In general, if we 

have a formula with a free variable x, then by adding “there is an x such 

that” in front of it we obtain a formula where x is not free anymore. 

In situations like that we say that “the variable x got bounded by the 

existential quantifier”. 

In a parallel way we may consider adding “for all x we have that” in 

front of a formula with a free variable x, asinthesentence 

• for all real numbers x we have that x 2 0. 

This operation also creates a formula where x is not free anymore 

and we say then that “the variable x got bounded by the universal 

quantifier”. 

Thus we have two 13 quantifiers: 

• the existential quantifier 9x (it is read “there is an x such that”, 

“for some x we have”), 

• the universal quantifier 8x (it is read “for all x we have”, “for 

every x we have”). 

12 unless there are some conventions around, see later 

13 we may consider more quantifiers but these two are the most important


When you add them in front of a formula with a free variable x, we 

make this variable bounded. If x was the only free variable, then (in a 

sense) we may ask if the formula obtained this way is true or false. 

Some examples of mathematical formulas with quantifiers: 

(⇤) 8 (8x)(x 2 0) 

(⇤) 9 (9x)(x 3 124x +3=0) 

(⇤) 10 (9y)(5 sin(y) =⇡). 

We may have multiple quantifiers: 

(⇤) 11 (8x)(8y)(x + y = y + x) 

(⇤) 12 (8x)(9y)(x + y =2013) 

(⇤) 13 (9x)(8y)(2 x > sin(y)) 

(⇤) 14 (8x)(8" >0)(9 >0)(8z)(|x z| < )|sin(x) sin(z)|

MATH 2230 — FALL 2013 17 

Warning: Never ever use quantifiers as abbreviations for English language 

phrases “for all” or “there exists” etc. Never write anything like 

“this solves the question 8x >0butwestillhavetodealwithc apple 0” 

or “we know that 9 a good object, but constructing it is more complicated”. 

3.2. Proofs Involving Quantifiers. What makes mathematical theorems 

complicated is a sequence of alternating quantifiers, like in the 

sentence in (⇤) 14 . However, even with only one quantifier around we 

may get lost if the arguments are not presented properly. 

Some sentences are true in every possible interpretation (model). 

For instance, (9x)(x >0) ) (9x)(x >0) is always true, regardless 

of our interpretation of >. All (suitable) substitutions to tautologies 

of propositional calculus will have this property, but not only those. 

In this part we will meet several sentences which are true regardless 

of their interpretation. Sentences of this type are called logical validities, 

orlogical axioms, ortautologies of the quantifier calculus. We 

will give some justifications for them – these arguments could serve as 

“templates” for all “common sense” proofs involving quantifiers. 

So, from now on let us fix: 

• the non-empty domain (universe) U of all variables, 

• formulas (x), (x) withatmostonefreevariablex. 

When a is an object from U, then we write (a) forthesentence 

obtained from (x) byreplacingeveryfreeoccurrenceofx with a. 

Similarly for . 

We will use x 0 to denote some fixed objects from U —note,thisis 

not a variable! 

Remark 3.2.1. Many mathematical theorems have the form of an implication. 

There are various strategies to prove the conditional ' ) but 

most often the direct approach is best. We assume that the hypothesis 

(antecedent) ' is true and we argue that the conclusion (consequent) 

holds true. 

Sometimes we want to argue for a biconditional statement ' , . 

Then most often we prove the two implications separately (remembering 

Corollary 2.2.5(9)). 

Proposition 3.2.2. The sentence 

(8x) (x) ) (9x) (x)


is a tautology of the quantifier calculus. 

Proof. Assume that (8x) (x) is true, that is, for every x 0 ,thesentence 

(x 0 )holdstrue. Chooseanyx 0 from the domain U and fix it. Then it 

follows that (x 0 )istrue. Consequently,x 0 witnesses that there exists 

an x such that (x) istrue. Thus,(9x) (x) holds,sowehaveshown 

the implication 

holds. This completes the proof. 

(8x) (x) ) (9x) (x) 

Proposition 3.2.3. The following sentences are tautologies of the quantifier 

calculus: 

(a) (8x)( (x) ^ (x)) , (8x) (x) ^ (8x) (x). 

(b) (9x)( (x) _ (x)) , (9x) (x) _ (9x) (x). 

(c) (8x) (x) _ (8x) (x) ) (8x)( (x) _ (x)). 

(d) (9x)( (x) ^ (x)) ) (9x) (x) ^ (9x) (x). 

Proof. (a) To prove this biconditional we will argue for the two implications 

separately. 

()) Let us assume that 

(i) (8x) (x) ^ (x) holds true. 

Suppose that x 0 is arbitrary but fixed. It follows from (i) that 

is true, and hence 

that 

(ii) 

(x 0 ) ^ (x 0 ) 

(x 0 )holdstrue. Sincex 0 was arbitrary we conclude 

(8x) (x) holdstrue. 

Again assuming x 0 is arbitrary, we know that (x 0 ) ^ (x 0 )istrueby 

(i). It follows that (x 0 ) is true as well. Hence, we get 

(iii) 

(8x) (x) holdstrue. 

Finally, clauses (ii) and (iii) imply that the conjunction 

(8x) (x) ^ (8x) (x) 

is true, completing the proof of the implication. 

(() Assume that (8x) (x) ^ (8x) (x) holdstrue.Thus 

and 

(⇤) 

(⇤⇤) 

(8x) (x) istrue, 

(8x) (x) istrue. 

⇤

MATH 2230 — FALL 2013 19 

Now, suppose that x 0 is arbitrary but fixed. It follows from (⇤) that 

(x 0 )istrue,andby(⇤⇤) wegetthat (x 0 )holds.Therefore, 

(x 0 ) ^ (x 0 ) 

is true. Since x 0 was arbitrary, we conclude that (8x)( (x) ^ (x)) is 

true. 

Consequently, the desired implication is true and the proof of the 

biconditional is complete. 

(b) To show this biconditional we will prove the two implications 

separately. 

()) Assume (9x)( (x) _ (x)) holds true. Let x 0 be such that 

(x 0 ) _ (x 0 ) 

is true. Hence, one of the statements, 

consider the two possible cases. 

(x 0 )or (x 0 ), is true. Let us 

Case 1: (x 0 )istrue. 

Then, x 0 witnesses that (9x) (x) holds true. Consequently, the disjunction 

(9x) (x) _ (9x) (x) 

holds true. 

Case 2: (x 0 )istrue. 

Then, similarly as in the previous case, except interchanging and , 

we conclude that the disjunction 

holds true. 

(9x) (x) _ (9x) (x) 

In both cases we concluded that the disjunction (9x) (x)_(9x) (x) 

holds true. Therefore the implication is justified. 

(() Assume (9x) (x) _ (9x) (x) holdstrue. Thus,either(9x) (x) 

or (9x) (x) istrueandthisleadstotwopossiblecases. 

Case 1: (9x) (x) istrue. 

Then we may choose an x 0 so that (x 0 )holds. Itthenfollowsthat 

(x 0 ) _ (x 0 )istrue.Consequently,x 0 witnesses that 

(9x)( (x) _ (x)) 

holds true. 


By interchanging and in the previous case, we conclude that 

(9x)( (x) _ (x))


holds true. 

In both cases we conclude that (9x)( (x) _ (x)) is true. Therefore 

the proof of the implication is complete. Hence, the biconditional (b) 

is shown. 

(c) Assume that (8x) (x)_(8x) (x) holds true. Then either (8x) (x) 

or (8x) (x) istrue,sowemayconsidertwopossiblecases. 


Let x 0 be arbitrary but fixed. By the assumption of the current case, 

(x 0 ) holds true. Hence, the disjunction (x 0 ) _ (x 0 )alsoholdstrue. 

Since x 0 was arbitrary, we conclude 

(8x)( (x) _ (x)) 

holds true. 


Interchanging and in the previous argument we justify that in the 

current case 

(8x)( (x) _ (x)) 

also holds true. 

In both possible cases we concluded that 

(8x)( (x) _ (x)) 

holds true, so the implication is shown. 

(d) Assume (9x)( (x) ^ (x)) holds true. Let x 0 be a witness for 

this, that is it is chosen so that (x 0 ) ^ (x 0 )istrue. Thenboth (x 0 ) 

and (x 0 )aretrue. Since (x 0 )istrue,weseethat(9x) (x) holds, 

and since (x 0 )istrueweseethat(9x) (x) holds.Consequently,the 

conjunction 

(9x) (x) ^9x (x) 

holds true, and this completes the proof of the implication. 

We have shown finally, that the four statements in Proposition 3.2.3 

are true, and this completes the proof. 

⇤ 

Remark 3.2.4. The sentence 

(8x)( (x) _ 

(x)) ) (8x) (x) _ (8x) (x) 

may be false. For instance, let x range over natural numbers and (x) 

means “x is even” and (x) means“x is odd.” Then, (8x)( (x)_ (x)) 

holds true, but both (8x) (x) and(8x) (x) arefalse.

MATH 2230 — FALL 2013 21 

Definition 3.2.5. Two sentences and (of quantifier calculus) are 

equivalent if the biconditional ( , ) is a tautology of the quantifier 

calculus. 

Formulas (x) and (x) areequivalent if the sentence 

(8x)( (x) , (x)) 

is a tautology. Similarly if more free variables are involved. 

Example 3.2.6. ¿Thefollowingpairsofformulasareequivalent: 

(1) (8x)( (x) ^ (x)) and (8x) (x) ^ (8x) (x). 

(2) (9x)( (x) _ (x)) and (9x) (x) _ (9x) (x). 

(3) ¬(9x) (x) and(8x)¬ (x). 

(4) ¬(8x) (x) and(9x)¬ (x). 

(5) ¬(8x)(9y) (x, y) and(9x)(8y)¬ (x, y). 

3.3. Syntax: First Order Logic. We start with fixing a lis of symbols 

that we will use in our formulas. Some symbols are independent 

of the area of mathematics that we are working in are called “logical” 

or “syntax” symbols. “Logical” symbols include: 

(~) q 1 variables: x, x 1 ,x 2 ,..., y, y 1 ,y 2 ,... and z,z 1 ,z 2 ,...; 

(~) q 2 grouping symbols: (, ) 

to increase readability we may use also [ , ], but they should be 

interpreted just as a di↵erent way of writing round parentheses; 

(~) q 3 logical connectives: ^, _, ¬, ), ,; 

(~) q 4 quantifiers: 9, 8. 

We also have area specific symbols which may include: 

(~) q 5 constants like 0, 1, 2,e,⇡ etc; 

(~) q 6 relational symbols like


1, while x y has an arity of 2. Often times, if F is a binary functional 

symbol, then we often write it between the terms. For example, 2 + 7, 

3 · x, x y, andsoon.OtherwiseweusetheformF (x, y, z) etc. 

Definition 3.3.1. (1) An alphabet for a first order language is a 

list A of 

(a) constants: c 1 ,c 2 ,c 3 ,... 

(b) relational symbols: R 1 ,R 2 ,R 3 ,... 

(each with fixed arity) and 

(c) functional symbols: F 1 ,F 2 ,F 3 ,... 

(each with fixed arity). 

(2) For the alphabet A, wedefineterms and formulas of the language 

L(A) asfollows. 

(a) Variables and constants are terms. 

(b) If F i is an n-ary function symbol and t 1 ,...,t n are terms, 

then F (t 1 ,...,t n )isaterm. 

(c) If t 1 ,t 2 are terms, then t 1 = t 2 is a formula. 

(d) If t 1 ,...,t n are terms and R is an n-ary relational symbol, 

then R(t 1 ,...,t n )isaformula. 

(e) If , ,... are formulas, then so are 

¬ , ( ^ ), ( _ ), ( ) ), ( , ). 

(f) If is a formula, then both (8x) and (9x) are formulas. 

Remark 3.3.2. If the function symbol is binary and we put it between 

the terms (i.e., we write t 1 Ft 2 ), then we add parentheses. So we write 

then (t 1 Ft 2 ). Without the parentheses, the structure of complex terms 

can be misinterpreted. 

Definition 3.3.3. Aformulawithoutfreevariablesiscalledasentence. 

Defining derivations in first order logic is somewhat more complicated 

than in propositional calculus (subsection 2.3). We may be lazy 

though and take as the logical axioms all tautologies of the quantifier 

calculus we can come with, including all substitutions to tautologies of 

propositional calculus. Then there are two rules of inference: modus 

ponens and the substitution rule. The latter can be described as follows 

15 : 

If t is a term and is a formula possibly containing the variable x, 

then [t/x] istheresultofreplacingallfreeinstancesofx by t in . 

The substitution rule states that for any and any term t, onecan 

conclude [t/x] from provided that no free variable of t becomes 

bound during the substitution process. 

15 quoted after Wikipedia

MATH 2230 — FALL 2013 23 

Thus we develop syntactics like in the propositional calculus case. We 

also study semantics (the model theory), but this is a totally di↵erent 

story. 

3.4. Exercises. 

Problem 3.1. Using binary relational symbols


Problem 3.4. Let (x, y) andbeaformulawithatmosttwofree 

variables x and y. Suppose also that no occurrence of x or y in is 

bounded by a quantifier. Prove that the following sentences are true. 

(1) (8x)(8y) (x, y) ) (8x) (x, x) 

(2) (9x) (x, x) ) (9x)(9y) (x, y) 

(3) (9x)(8y) (x, y) ) (8y)(9x) (x, y) 

Problem 3.5. Justify Remark 3.2.6. 

Problem 3.6. Which of the following sentences are always true (i.e., 

which sentences are tautologies of the quantifiers calculus) ? Justify 

those which are tautologies, provide counterexamples for those which 

are not. 

(1) (8x) (x, x) ) (8x)(8y) (x, y) 

(2) (9x)(9y) (x, y) ) (9x) (x, x) 

(3) (8y)(9x) (x, y) ) (9x)(8y) (x, y) 

(4) (8y)(8x) (x, y) , (8x)(8y) (x, y) 

(5) (9y)(9x) (x, y) , (9x)(9y) (x, y) 

(6) (8x) (9x) (x) ) (x) 

(7) (8x) (x) ) (9x) (x) 

(8) (8x) (x) ) (8x) (x) ) 8x (x) ) (x) 

(9) (8x) (x) , (8x) (x) ) 8x (x) , (x) 

(10) (8x) (x) ) (8y) (y) 

Problem 3.7. Aformulaisinprenex form if it is written as a string 

of quantifiers (referred to as the prefix) followed by a quantifier–free 

part (referred to as the matrix). 16 For each of the following formulas 

find a formula in the prenex form equivalent to the given formula and 

the matrix simplified as much as possible. 

(1) ¬(8x)(8y)(x + y = y + x) 

(2) ¬(8x)(9y)(x + y =2013) 

(3) ¬(9x)(8y)(2 x > sin(y)) 

(4) ¬(8" >0)(9 >0)(8x)(8z)(|x z| < )|sin(x) sin(z)| 0)(9 >0)(8z)(|x z| < )|2 x 2 z |

Quantifiers - University of Nebraska Omaha

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