Logic as a tool: Quantification David Gries Professor Emeritus Computer Science, Cornell

Logic as a tool: QuantificationDavid GriesProfessor EmeritusComputer Science, Cornell Based on an in-progress revision of A Logical Approach to Discrete Math by Gries and Schneider

Note: This is a workshop, and not a full-fledged class. We touch on the highlights, but the slides contain a lot more informationWhen I teach a course on discrete structures, I spend:1. 6-7 1-hour lectures on propositional logic2. 6 1-hour lectures on quantification (predicate logic)Between lectures, students have homework. They end up proving 50-60 theorems of logic and gain real understanding and a skill in syntactic calculation. One could call it fluency in developing proofs.

Quantification • Provide a single uniform notation for all quantifications, unifying what has been confusion. Scope, dummy, bound variable, free or bound occurrence of a variable, etc. can be discussed in a single setting • i=19 i*i • i . 0 ≠ i i2 > 0 (for all) •  i . 0 ≠ i i2> 0 (exists) • Provide axioms and theorems for manipulating quantifications. This will give you basic information, which has been missing from your mathematics education until now

Notation for Quantification • Mathematical convention Our notation • i=19 i*i (+ i | 1 ≤ i ≤ 9: i*i) • i=19 i*i (* i | 1 ≤ i≤ 9: i*i) • i . 0 ≠ i i2 > 0 (i | 0 ≠ i: i2> 0) • i . 0 ≠ i i2 > 0 (i | 0 ≠ i: i2 > 0) operator dummy range body Dummy sometimes called the bound variable

How to evaluate (+ i | 1 ≤ i ≤ 4: i*i) • 1. Write down values iin the range 1 ≤ i ≤ 4: • 1, 2, 3, 4 • 2. Write down the values of the body,i*i, one for each possible value of i in the range: • 1, 4, 9, 16 • 3. Put the operator + between each pair of values: • 1 + 4 + 9 + 16 • 4. Evaluate the expression: 30

How to evaluate (i | –1 ≤i ≤ 2: i2 > 0) • 1. Write down values iin the range –1 ≤ i ≤ 2: • -1, 0, 1, 2 • 2. Write down the values of the body, i2 > 0, one for each possible value of i in the range: • 1> 0, 0 > 0, 1> 0, 2 > 0 • 3. Put the operator between each pair of values: • 1 > 0 0 > 0  1 > 0  2 > 0 • 4. Evaluate the expression: false

What operators are allowed in a quantification? • (oi | range : body) • Operator ocan be any operator that is associative and symmetric. If a range of false comes into play, it also needs an Identity. • (gcdi | 1 ≤ i ≤ 9: i*i) (greatest common divisor) • (i | 1 ≤ i ≤ 9: i*i) (minimum) • (i| 1 ≤ i ≤ 9: I > 0) (equivales)

General form of Quantification o: operator i: dummy (bound variable) t: type of dummy (often omitted) R: range E: body • (oi:t | range : E) • Value: • E[i:= v0] oE[i:= v1] oE[i:= v2] o… • where {v0, v1, v2, …} is the set of valuesfor which R is true. • (oi |: E)is abbreviation for (o i | true: E) • Examples of quantifications with types • (+i:int| 1 ≤ i ≤ 2: i*i) = 1 + 4 • (+i:real| 1 ≤ i ≤ 2: i*i) isundefined

Scope, bound variables, free variables • (oi:t | R: P)Scope of dummy i is R and P • Scope rules for dummy similar to rules for local variable of a procedure • publicvoid p(int x) {intk; k= x+x; k= 2; } • Scope of local variable kis procedure body. Any occurrence • of koutside procedure body refers to entirely different entity, which just happens to have same name. In same way, scope of i in • (oi| R: P) • is R and P. An occurrence of ioutside this expression refers to an entirely different entity

Scope, bound variables, free variables • Occurrence of vis bound in an expiff it is in the scope of dummy v. It is bound to nearest such dummy. • Occurrence of v is freeif it is not bound. • occurs(“v”, “E”) means that at least one variable in list v of variables occurs free in at least one expin explist E. • Occurrences of a var in an expwithout quantifications are free • i > 0  (i | 0 ≤ i: x * i = 0) Bound to dummy i Free. Gets value from state

Translating English into Quantifications • Canonical English phrase for a quantification indicates operator, names dummies, makes explicit range and body • Sum over k of k*k for k in range 1..10 (+k | 1 ≤ k ≤ 10: k * k) • Product over j of 2*j for even integers j between 0 and 10 • (*j | even.j and 0 ≤ j ≤ 10: 2*j) • Minimum value over all j of j3 – j2(j | true: j3 – j2) • Conjunction of even.jfor positive j (j | 0 < j: even.j) • For every odd integer k, 3*k is odd (k | odd.k: odd(3*k)) • There exist two positive integers i and j whose sum is negative • (i,j| 0 < i0 < j: i + j < 0)

It helps to put phrases in canonical form • Example • sum of the square of the odd positive integers that are less than 50. • Dummy is not explicit so reword • sum of i*i for i an odd positive integer that • is less than 50 • Nowformalize • (+ i | odd.i0 < i ≤ 50: i*i)

Universal Quantification • is associative and symmetric; can use in quantification • (i| 1 ≤ i < 4: i < i2) (i | 1 ≤ i < 4: i < i2) • is equivalent to • 1 < 12  2< 22  3< 32(false) • so it is true iff all conjuncts are true. Hence, read as • For all i such that i is in 1..3, i is less than i squared • Other words that signal universal quantification: • Every all for all for each any

Formalizing English statements • Strategy: If a phrase is to be formalized as universal quantification, first put it in canonical form • For all i, [describe range of i], … holds • Sentence: x is less than every positive integerCanonical form: For all i, i a positive integer, x < iFormalization: (i | i > 0: x < i) • Sentence: Even integers are redCanonical form: For all integers k, where k is even, k is redFormalization: (k | even.k: k is red) • Sentence: Paragraphs have at least two sentencesCanonical form: For every paragraph p, p has at least two sentencesFormalization:(p:para : p has at least two sentences)

Convention • Convention: Free variables in a theorem are considered to be implicitly universally quantified • Example: red.x • If no particular state is implied by the discussion, the meaning is that every value xsatisfies red.x, i.e. ( x |: red.x) • Example: Red numbers are pretty

Existential Quantification • is associative and symmetric; can use in quantification • ( i| 1 ≤ i < 4: i < i2) (i| 1 ≤ i < 4: i2) • is equivalent to • 1 < 12  2< 22  3< 32(false) • so it is true (at least) if one disjunct is true —if there exists an integer i in range 1..3 such that i2 • Existential quantifications are signaled by • exists some there are there is at least one

Formalizing English statements • Strategy: If a phrase is to be formalized as an existential quantification, put it in canonical form: • There exist i, [describe range of i], such that …holds • Sentence: Some even integer is greenCanonical form: There is an i that is even such that i is greenFormalization: ( i| even.i: i is red) • Sentence: At least one chapter has an even number of pagesCanonicalform:There exists a chapter c such that c has an even number of pagesFormalization: ( i:chap |: even(c.size))

Negation and Quantification • What is negation of All integers are even • Not: All integers are not evenbut: Not (all integers are even)which is: Some integer is not even • Not (all integers are even) • = <Formalize using quantification> • !( z |: even.z) • = <Rewrite as a long conjunction> • !(…  even(–1)  even.0  even.1  even.2  …)= <De Morgan (9.23c)> • (…  !even(–1)  !even.0  !even.1  ! even.2  …) • = <Rewrite using using quantification> • ( z |: ! even.z) • = <Return to English> • Some integer is not even

Take care with formalizations • Define: mother(m, c): “m is c’s mother” • English: Every child has a motherCanonical form: For every child c, there is a mother for cFormalization: (c |: (m |: mother(m, c))) • English: There is a mother for every childCanonical form: There exists a mother m such that for every child c, m is c’s motherFormalization: ( m|: ( c|: mother(m, c))) • Means that there is one mother who is the mother of all! Not whatis meant! Probably meant: Every child has a mother

Divide and Conquer • Sentence too long/complicated to formalize in one step? Do it in several steps • English: An even integer is smaller than some odd integerCanonical form: For every even integer i, iis smaller than some odd integer • Formalize: (i| even.i: i is smaller than some odd integer) • Work on body: iis smaller than some odd integerCanonical form: There exists an odd integer j such that i < jFormalize: ( j| odd.j: i < j) • Final form: (i| even.i: ( j| odd.j: i < j))

Contextual Substitution Revisited • (o x | R: P) E[v:= P] • requires different definition because quantifications introduce local variables. • Contextual substitution in a quantification: • If y does not occur free in v or F • (oy | R: P)[v:= F] = (o y | R[v:= F]: P[v:= F]) • If y does occur free in v or F, let z be a fresh variable (it does not occur free in v, R, P, or F). Then • (o y | R: P)[v:= F] = (oz | R[y:= z][v:= F] : P[y:= z][v:= F])

Reason for changing dummy • (+i | i = n: i) = n • Therefore (+i | i = n: i)[n:= i] = n[n:= i] = i • But also, without required change in dummy, we have • (+i | i = n: i)[n:= i] = (+i | i = i: i) —bad substitution • Latter is undefined because all values ofisatisfy i = i • Correct way • (+i | i = n: i)[n:= i] • = <Contextual substitution> • (+j| j= n: j)[n:= i] • = <Definition of quantification> • i

Pure Predicate Logic • Predicate logic formula: boolean expression in which some boolean variables may have been replaced by • Predicates: calls on booleanfunctions whose arguments may be of types other than booleanExampleequal(x, x – z + z)Exampleless(x, y + z) Function names are called predicate symbols Arguments are called terms • Universal and existential quantificationExamplex < y x = z  q(x, z + x) • Predicates: x < y, x = z, q(x, z + x)Terms: x, y, z, z + x

Pure Predicate Logic • Axioms: Axioms of propositional logic Axioms for (x | R:P) —we will give more Axioms for ( x | R: P) —we will give more • Inference rulesLeibniz (3.1) • Transitivity (3.2) • Equanimity (3.3 • Leibniz for quantification (8.18)

Theories • In pure predicate logic, function symbols are uninterpreted(except for equality =). The logic provides no rules for manipulating them. We will have general rules for manipulation that are sound no matter what meanings are given to function symbols • To get a theory: add axioms that give meaning to someuninterpretedfunction symbols • Example: Theory of integers: pure predicate logic plus axioms for manipulating operators(i.e. functions +, –, <, <=, etc.) • Example: Theory of sets: add axioms for manipulating set expressions containing operators like membership , union  • Example:Joint theory of sets and integers allows reasoning about expressions that contain both.

Theorems for Quantification • Theorems hold only if each quantification is well-defined and the operator has identity u. • (+ i| 2 ≤ i ≤ 3: i*i) = 4 + 9(+ i | 2 ≤ i ≤ 2: i*i) = 4 • (+ i | 2 ≤ i ≤ 1: i*i) = ? Answer is 0, becausethe Identity of + is 0: 0 + x = x (* i | 2 ≤ i ≤ 3: i*i) = 4 * 9 (* i | 2 ≤ i ≤ 2: i*i) = 4 (* i | 2 ≤ i ≤ 1: i*i) = ? Answer is 1, becausethe Identity of * is 1: 1 * x = x

Axioms for Quantification • (8.22) Axiom One-point rule: Provided occurs(‘x’, ‘E’),(o x | x = E: P) = P[x:= E] • Why the caveat? Consider this: • (+ i|i = i+ 1 : i) = 0 (since range is empty) • Butif no caveat, • (+ i |i = i+ 1 : i) = i[i:= i+1], which is i+1 • Consider (o x | x = E: P) = P[x:= E] where x occurs free in E • LHS has no free occurrence of x.RHS has free occurrence of x.How could they be equal in all states, in general?

Axioms for Quantification • (8.23) Axiom Distributivity: Provided P,Q: bool or R is finite • (o x | R: P) o (o x | R: Q) = (o x | R: P oQ) • Example: • values of P for x values of Q for x satisfying R satisfying R 2 + 5 + 8 3 + 1 + 9 • 2 + 3 + 5 + 1 + 8 + 9 Rest on fact that operator ois associative and symmetric

Axioms for Quantification • (8.30) Change of dummy: Provided occurs(‘y’, ‘R, P’) and f has inverse • (ox | R: P) = (oy | R[x:= f.y]: P[x:= f.y]) • (8.31) Dummy renaming: Provided occurs(‘y’, ‘R, P’)(o x | R: P) = (o y | R[x:= y]: P[x:= y]) • (8.32) Split off term (example): (o x | 0 ≤ i < n+1: P) = (o x | 0 ≤ i < n: P) o P[i:= n]

Use of Change of Dummy(o x | R: P) = (o y | R[x:=f.y]: P[x:= f.y]) • (+ x | 2 ≤ x ≤ 9: x2) = (+ y | 0 ≤ y ≤ 7: (y+2)2) • Convince yourself that this should hold Use function f.y = y + 2 f has an inverse R is 2 ≤ x ≤ 9 R[x:= f.y] is 2 ≤ y + 2 ≤ 9 which equals 0 ≤ y ≤ 7 P isx2 P[x:= f.y] is (y+2)2

Proof of Change of Dummy • The next slide contains a proof of Change of Dummy. • It is an opportunity driven proof: at each step, the shapes of the current formula and the goal give inspiration for the step. This makes the proof memorable. • Evidence that this proof is memorable comes from our courses. The students were told they would have to present this proof on a quiz. About 90 percent (out of 80–90 students) get it completely right. We believe that this would be impossible with an informal English proof.

= <One-point rule (8.22) —need quantification over x> (o y | R[x:= f.y]: (o x | x = f.y: P) • (o y | R[x:= f.y]: P[x:= f.y]) start with side with structure Head for (o x | R: P) = <Nesting (8.26) —move x to outside> (ox,y | R[x:= f.y] x = f.y: P) = <Substitution (3.91) —to get ready to remove R[x ..]> (ox,y | R[x:= x] x = f.y: P) = <R[x:= x] = R; Nesting (8.26) —move y inside> (o x | R: (o y | x = f.y: P) = <x = f.y y = f-1.x –prepare to eliminate y> (o x | R: (o y | y = f-1.x: P) = <One-point rule (8.22)> (o x | R: P[y = f-1.x]) = <Textual substitution –occurs(‘y’, ‘P’)> (o x | R: P)

Universal Quantification Theorems • is symmetric, is associative, and has an identity true • Write (x | R: P) or (x | R: P) • All axioms for quantification hold. • Additional axiom • (9.3) Axiom over : Provided occurs(‘x’, ‘P’),P (x | R: Q) ( x | R: P Q) • Reason for caveat: ensure that LHS and RHShave same set of free variables • Look at theorems at end for theorems for 

Trading • Trading: Moving a conjunct of the range to the body or vice versa. Here are the basic Trading theorems • (9.4) Trading: (x | R: P)  (x |: R  P) • (9.5a) Trading: ( x | Q R: P) ( x | Q: R  P)

Modify Metatheorem Monotonicity • The range of a quantification is an antimonotonicposition. So we have • (4.7) Metatheorem Monotonicity (see theorems) • Example of use • ( i | i = 3: i*I > 0) •  <Antimonotonicity: i = 3  i > 0> • ( i| i > 0: i*I > 0) i = 3 appears in 1 (an odd number) antimonotonic position. So the direction of the arrow changes.

Instantiation • (9.17) Instantiation: ( x |: P)  P[x:= E] • One-point rule sharper than Instantiation. In some situations Instantiation is useful. Like symmetry, often used implicitly. • Implicit use of Instantiation more concealed if universal quantification itself is not written formally. • Example Conventionally write • (9.20) ( a,b |: a + b = b + a) • as • a + b = b + a for all integers a, b • Because universal quantification is a side comment and not part of the formula, it is easy to forget that producing, sayx*y + z = z + x* y • From (9.20) requires Instantiation

Metatheorem • (9.16) Metatheorem. P is a theorem iff ( x |: P) is a theorem. • Use this Metatheorem to discuss ways of stating axioms and theorems in mathematics. Below are three different ways: • Axiom: ( b, c: int |: b + c = c + b) • Axiom: b + c = c + b for integers b, c • In this chapter, band c have type integerAxiom: b + c = c + b

Existential Quantification • is symmetricand associative and has a identity true • (x | R: P) or (x | R: P) • All axioms for quantification hold for . • Additional axiom: • Axiom Generalized De Morgan:( x | R: P)  ( x | R: P) • Idea behind generalizationP1  P2  (P1 P2) • Generate theorems for from theorems for  • Use Unfold-fold to prove them.

Opportunity-driven proof • Prove (9.40), but with all ranges true: • ( x |: (y |: P))  (y|: (x |: P)) • There is a reason for each step, which is discussed in the hint. • ( x |: (y |: P)) = <Quantified freshy (9.9) —Universal quantification over must appear at the outer level this gets it there.> (y |: ( x |: (y |: P)))  <Monotonicity: Instantiation (9.17) —The inner quantification over y has to be removed This does it> (y |: ( x |: P))

Using a Witness • New proof method to prove Interchange of quantifications (9.40) with ranges true: • ( x |: (y |: P))  (y |: ( x |: P)) • The antecedent indicates there exists a value for dummy x for which (y |: P) holds. Give the value a namexxand use instead the antecedent (y |: P)[x:= xx].Thus prove the theorem by proving (y |: P)[x:= xx] (y |: ( x |: P)) • (y |: P[x:= xx]) • = <One-point rule (8.22)> • (y |: ( x | x = xx: P)) • <Monotonicity: x =xx  true> • (y |: ( x |: P))

Proving an Argument Sound soc: Socratesh.p: person p is humanm.p: person p is mortal • Socrates is human. • All humans are mortal. • Therefore some person is mortal. h.soc (p |: h.pm.p)  ( p |: m.p) h.soc (p |: h.pm.p) (heading for ( p |: m.p)  <Monotonicity: Instantiation (9.13), with p:= soc> h.soc (h.soc m.soc)  <Modus ponens (3.77)> m.soc  <-Introduction (9.28)> ( p |: m.p)

Proving an Argument Sound • All mosquitos are pests. No butterflies are pests. Therefore, butterflies are not mosquitos. • (c|: m.cp.c)  (c|: b.c p.c) (c| b.c: p.c) • Motivate first step. Must weaken antecedent a  quantification over c. First conjunct of antecedent already has this form. Application of De Morgan to second puts second in that form. • (c|: m.cp.c)  (c|: b.c p.c)= <DeMorgan (9.23b); DeMOrgan (3.52a)>(c|: m.cp.c)  (c|: b.cp.c)= <Distributivity (8.23)>(c: (m.cp.c)  (b.cp.c))=<Implication (3.65)>(c: (m.cp.c)  (p.cb.c)) —you finish it

Mathematical Reasoning • Is there some real xfor which 1/(x*x + 1) > 1? • Formalized is following valid? (x:real |: 1/(x*x + 1) > 1) (x|: 1/(x*x + 1) > 1)= <Arithmetic>(x|: 1 > x*x + 1)= <Arithmetic>(x|: 0 > x*x)=<x*x ≥ 0 —from theory of the reals>(x|: false)= <(9.28)>false

Set Theory • View set theory as an extension of predicate logic. Add a new type, notations for constants and operations, and axioms • Usual notations: • {x | P} set of values of x that satisfy predicate P • {x*x| 0 ≤ x ≤ 100} • {x*y | 0 ≤ x*y ≤ 100} what’s the dummy? • Our notation: • {x | P: x} • {x | 0 ≤ x ≤ 100: x*x} • {x |0 ≤ x*y ≤ 100: x*y}

Comprehension and enumeration • {x | P | E} the set of values of expression E where dummy x ranges over values that satisfy P • {v0, v1, …, vn} shorthand for {x | x = v0  x = v1  …  x = vn: x} • {1, 3, 1} = {x | x = 1  x = 3  x = 1: x} = {1, 3}

Membership and Extensionality • (10.4) Axiom Set membership: Provided not occurs(‘x’, ‘F’), • F  {x | R: E}  (x | R: F = E} • (10.5) Axiom Extensionality: S = T  (x |: x S  x T} • Theorems • e{x| false: E}  false • {x| false: E}  {} • v  {x| x = e}  v = e • S = {x | x S: x}

Logic as a tool: Quantification David Gries Professor Emeritus Computer Science, Cornell