Computation Theory. 12 lectures for University of Cambridge 2015 Computer Science Tripos, Part IB by Prof. Andrew Pitts.

Transcription

1 Computation Theory 12 lectures for University of Cambridge 2015 Computer Science Tripos, Part IB by Prof. Andrew Pitts c 2015 AM Pitts

2

3 Contents Introduction: algorithmically undecidable problems 6 Decision problems. The informal notion of algorithm, or effective procedure. Examples of algorithmically undecidable problems. [1 lecture] Register machines 18 Definition and examples; graphical notation. Register machine computable functions. Doing arithmetic with register machines. [1 lecture] Coding programs as numbers 34 Natural number encoding of pairs and lists. Coding register machine programs as numbers. [1 lecture] Universal register machine 41 Specification and implementation of a universal register machine. [1 lecture] The halting problem 50 Statement and proof of undecidability. Example of an uncomputable partial function. Decidable sets of numbers; examples of undecidable sets of numbers. [1 lecture] Turing machines 59 Informal description. Definition and examples. Turing computable functions. Equivalence of register machine computability and Turing computability. The Church-Turing Thesis. [2 lectures] Primitive and partial recursive functions 84 Definition and examples. Existence of a recursive, but not primitive recursive function. A partial function is partial recursive if and only if it is computable. [2 lectures] Lambda calculus 103 Alpha and beta conversion. Normalization. Encoding data. Writing recursive functions in the λ-calculus. The relationship between computable functions and λ-definable functions. [3 lectures]

4 These notes are designed to accompany 12 lectures on computation theory for Part IB of the Computer Science Tripos. The aim of this course is to introduce several apparently different formalisations of the informal notion of algorithm; to show that they are equivalent; and to use them to demonstrate that there are uncomputable functions and algorithmically undecidable problems. At the end of the course you should: be familiar with the register machine, Turing machine and λ-calculus models of computability; understand the notion of coding programs as data, and of a universal machine; be able to use diagonalisation to prove the undecidability of thehaltingproblem; understand the mathematical notion of partial recursive function and its relationship to computability. The prerequisites for taking this course are the Part IA courses Discrete Mathematics and Regular Languages and Finite Automata. This incaration of the Computation Theory course builds on previous lecture notes by Ken Moody, Glynn Winskel, Larry Paulson and myself. It contains some material that everyone who calls themselves a computer scientist should know. ItisalsoaprerequisiteforthePartIBcourseon Complexity Theory. An exercise sheet, any corrections to the notes and additional material can be found on the course web page. 4 Recommended books Hopcroft, J.E., Motwani, R. & Ullman, J.D. (2001). Introduction to Automata Theory, Languages and Computation, Second Edition. Addison-Wesley. Hindley, J.R. & Seldin, J.P. (2008). Lambda-Calculus and Combinators, an Introduction. Cambridge University Press (2nd ed.). Cutland, N.J. (1980) Computability. An introduction to recursive function theory. Cambridge University Press. Davis, M.D., Sigal, R. & Wyuker E.J. (1994). Computability, Complexity and Languages, 2nd edition. Academic Press. Sudkamp, T.A. (1995). Languages and Machines, 2ndedition. Addison-Wesley. 5

5 Introduction 6

6 Algorithmically undecidable problems Computers cannot solve all mathematical problems, even if they are given unlimited time and working space. Three famous examples of computationally unsolvable problems are sketched in this lecture. Hilbert s Entscheidungsproblem The Halting Problem Hilbert s 10th Problem. 7 Hilbert s Entscheidungsproblem Is there an algorithm which when fed any statement in the formal language of first-order arithmetic, determines in a finite number of steps whether or not the statement is provable from Peano s axioms for arithmetic, using the usual rules of first-order logic? Such an algorithm would be useful! For example, by running it on k > 1 p, q (2k = p + q prime(p) prime(q)) (where prime(p) is a suitable arithmetic statement that p is a prime number) we could solve Goldbach s Conjecture ( every even integer strictly greater than two is the sum of two primes ), a famous open problem in number theory. 8

7 Hilbert s Entscheidungsproblem Is there an algorithm which when fed any statement in the formal language of first-order arithmetic, determines in a finite number of steps whether or not the statement is provable from Peano s axioms for arithmetic, using the usual rules of first-order logic? Posed by Hilbert at the 1928 International Congress of Mathematicians. The problem was actually stated in a more ambitious form, with a more powerful formal system in place of first-order logic. In 1928, Hilbert believed that such an algorithm could be found. AfewyearslaterhewasprovedwrongbytheworkofChurchand Turing in 1935/36, as we will see. 8 Decision problems Entscheidungsproblem means decision problem. Given asets whose elements are finite data structures of some kind (e.g. formulas of first-order arithmetic) apropertyp of elements of S (e.g. property of a formula that it has a proof) the associated decision problem is: find an algorithm which terminates with result 0 or 1 when fed an element s S and yields result 1 when fed s if and only if s has property P. 9

8 Algorithms, informally No precise definition of algorithm at the time Hilbert posed the Entscheidungsproblem, justexamples,suchas: Procedure for multiplying numbers in decimal place notation. Procedure for extracting square roots to any desired accuracy. Euclid s algorithm for finding highest common factors. 10 Algorithms, informally No precise definition of algorithm at the time Hilbert posed the Entscheidungsproblem, justexamples. Common features of the examples: finite description of the procedure in terms of elementary operations deterministic (next step uniquely determined if there is one) procedure may not terminate on some input data, but we can recognize when it does terminate and what the result is. 10

9 Algorithms, informally No precise definition of algorithm at the time Hilbert posed the Entscheidungsproblem, justexamples. In 1935/36 Turing in Cambridge and Church in Princeton independently gave negative solutions to Hilbert s Entscheidungsproblem. First step: give a precise, mathematical definition of algorithm. (Turing: Turing Machines;Church:lambda-calculus.) Then one can regard algorithms as data on which algorithms can act and reduce the problem to The Halting Problem is the decision problem with set S consists of all pairs (A, D), wherea is an algorithm and D is a datum on which it is designed to operate; property P holds for (A, D) if algorithm A when applied to datum D eventually produces a result (that is, eventually halts we write A(D) to indicate this). Turing and Church s work shows that the Halting Problem is undecidable, thatis,thereisnoalgorithmh such that for all (A, D) S { 1 if A(D) H(A, D) = 0 otherwise. 11

10 There s no H such that H(A, D) = { 1 if A(D) 0 otherwise. for all (A, D). Informal proof, bycontradiction.ifthereweresuchanh, let C be the algorithm: input A; compute H(A, A); if H(A, A) =0 then return 1, else loop forever. So A (C(A) H(A, A) =0) (since H is total) and A (H(A, A) =0 A(A) ) (definition of H). So A (C(A) A(A) ). Taking A to be C, we get C(C) C(C), contradiction! 12 From HP to Entscheidungsproblem Final step in Turing/Church proof of undecidability of the Entscheidungsproblem: theyconstructedanalgorithm encoding instances (A, D) of the Halting Problem as arithmetic statements Φ A,D with the property Φ A,D is provable A(D) Thus any algorithm deciding provability of arithmetic statements could be used to decide the Halting Problem so no such exists. 13

11 Hilbert s Entscheidungsproblem Is there an algorithm which when fed any statement in the formal language of first-order arithmetic, determines in a finite number of steps whether or not the statement is provable from Peano s axioms for arithmetic, using the usual rules of first-order logic? With hindsight, a positive solution to the Entscheidungsproblem would be too good to be true. However, the algorithmic unsolvability of some decision problems is much more surprising. A famous example of this is Hilbert s 10th Problem Give an algorithm which, when started with any Diophantine equation, determines in a finite number of operations whether or not there are natural numbers satisfying the equation. One of a number of important open problems listed by Hilbert at the International Congress of Mathematicians in

12 Diophantine equations p(x 1,...,x n )=q(x 1,...,x n ) where p and q are polynomials in unknowns x 1,...,x n with coefficients from N = {0, 1, 2,...}. Named after Diophantus of Alexandria (c. 250AD). Example: find three whole numbers x 1, x 2 and x 3 such that the product of any two added to the third is a square [Diophantus Arithmetica, BookIII, Problem7]. In modern notation: find x 1, x 2, x 3 N for which there exists x, y, z N with (x 1 x 2 + x 3 x 2 ) 2 +(x 2 x 3 + x 1 y 2 ) 2 +(x 3 x 1 + x 2 z 2 ) 2 = 0 16 Hilbert s 10th Problem Give an algorithm which, when started with any Diophantine equation, determines in a finite number of operations whether or not there are natural numbers satisfying the equation. Posed in 1900, but only solved in 1970: Y Matijasevič, J Robinson,M Davis and H Putnam show it undecidable by reduction from the Halting Problem. Original proof used Turing machines. Later, simpler proof [JP Jones & Y Matijasevič, J. Symb. Logic 49(1984)] used Minsky and Lambek s register machines we will use them in this course to begin with and return to Turing and Church s formulations of the notion of algorithm later. 17

13 Register machines 18

14 Register Machines, informally They operate on natural numbers N = {0, 1, 2,...} stored in (idealized) registers using the following elementary operations : add 1 to the contents of a register test whether the contents of a register is 0 subtract 1 from the contents of a register if it is non-zero jumps ( goto ) conditionals ( if_then_else_ ) 20 Definition. Aregistermachineisspecifiedby: finitely many registers R 0, R 1,...,R n (each capable of storing a natural number); aprogramconsistingofafinitelistofinstructionsof the form label : body, wherefori = 0, 1, 2,..., the (i + 1) th instruction has label L i. Instruction body takes one of three forms: R + L R L, L HALT add 1 to contents of register R and jump to instruction labelled L if contents of R is > 0, thensubtract 1 from it and jump to L,elsejumpto L stop executing instructions 21

15 registers: R 0 R 1 R 2 program: L 0 : R 1 L 1, L 2 L 1 : R + 0 L 0 L 2 : R 2 L 3, L 4 L 3 : R + 0 L 2 L 4 : HALT Example example computation: L i R 0 R 1 R Register machine computation Register machine configuration: c =(l, r 0,...,r n ) where l =currentlabelandr i =currentcontentsofr i. Notation: R i = x [in configuration c] means c =(l, r 0,...,r n ) with r i = x. Initial configurations: c 0 =(0, r 0,...,r n ) where r i =initialcontentsofregisterr i. 23

16 Register machine computation A computation of a RM is a (finite or infinite) sequence of configurations c 0, c 1, c 2,... where c 0 =(0, r 0,...,r n ) is an initial configuration each c =(l, r 0,...,r n ) in the sequence determines the next configuration in the sequence (if any) by carrying out the program instruction labelled L l with registers containing r 0,...,r n. 24 Halting For a finite computation c 0, c 1,...,c m,thelast configuration c m =(l, r,...) is a halting configuration, i.e. instruction labelled L l is either HALT (a proper halt ) or R + L, orr L, L with R > 0, or R L, L with R = 0 and there is no instruction labelled L in the program (an erroneous halt ) N.B. can always modify programs (without affecting their computations) to turn all erroneous halts into proper halts by adding extra HALT instructions to the list with appropriate labels. 25

17 Halting For a finite computation c 0, c 1,...,c m,thelast configuration c m =(l, r,...) is a halting configuration. Note that computations may never halt. For example, L 0 : R + 0 L 0 L 1 : HALT only has infinite computation sequences (0, r), (0, r + 1), (0, r + 2), Graphical representation one node in the graph for each instruction arcs represent jumps between instructions lose sequential ordering of instructions so need to indicate initial instruction with START. instruction representation R + L R + [L] [L] R L, L R [L ] HALT HALT L 0 START [L 0 ] 26

18 registers: R 0 R 1 R 2 program: L 0 : R 1 L 1, L 2 L 1 : R + 0 L 0 L 2 : R 2 L 3, L 4 L 3 : R + 0 L 2 L 4 : HALT Example graphical representation: START R 1 R + 0 R 2 R + 0 HALT Claim: starting from initial configuration (0,0, x, y), this machine s computation halts with configuration (4, x + y,0,0). 27 Partial functions Register machine computation is deterministic: in any non-halting configuration, the next configuration is uniquely determined by the program. So the relation between initial and final register contents defined by a register machine program is a partial function... Definition. A partial function from a set X to a set Y is specified by any subset f X Y satisfying (x, y) f (x, y ) f y = y for all x X and y, y Y. 28

19 Partial functions ordered pairs {(x, y) x X y Y} i.e. for all x X there is at most one y Y with (x, y) f Definition. ApartialfunctionfromasetX to a set Y is specified by any subset f X Y satisfying (x, y) f (x, y ) f y = y for all x X and y, y Y. 28 Notation: Partial functions f(x) =y means(x, y) f f(x) means y Y ( f(x) =y) f(x) means y Y ( f(x) =y) X Y =setofallpartialfunctionsfromx to Y X Y =setofall(total) functionsfromx to Y Definition. ApartialfunctionfromasetX to a set Y is total if it satisfies f (x) for all x X. 28

20 Computable functions Definition. f N n N is (register machine) computable if there is a register machine M with at least n + 1 registers R 0, R 1,...,R n (and maybe more) such that for all (x 1,...,x n ) N n and all y N, the computation of M starting with R 0 = 0, R 1 = x 1,...,R n = x n and all other registers set to 0, haltswithr 0 = y if and only if f (x 1,...,x n )=y. Note the [somewhat arbitrary] I/O convention: in the initial configuration registers R 1,...,R n store the function s arguments (with all others zeroed); and in the halting configuration register R 0 stores it s value (if any). 29 registers: R 0 R 1 R 2 program: L 0 : R 1 L 1, L 2 L 1 : R + 0 L 0 L 2 : R 2 L 3, L 4 L 3 : R + 0 L 2 L 4 : HALT Example graphical representation: START R 1 R + 0 R 2 R + 0 HALT Claim: starting from initial configuration (0,0, x, y), this machine s computation halts with configuration (4, x + y,0,0). So f (x, y) x + y is computable. 30

21 Computable functions Recall: Definition. f N n N is (register machine) computable if there is a register machine M with at least n + 1 registers R 0, R 1,...,R n (and maybe more) such that for all (x 1,...,x n ) N n and all y N, the computation of M starting with R 0 = 0, R 1 = x 1,...,R n = x n and all other registers set to 0, haltswithr 0 = y if and only if f (x 1,...,x n )=y. 31 Multiplication f (x, y) xy is computable START R + 3 R 1 R 2 R + 0 HALT R 3 R + 2 If the machine is started with (R 0, R 1, R 2, R 3 )=(0, x, y,0), ithalts with (R 0, R 1, R 2, R 3 )=(xy,0,y,0). 32

22 Further examples The following arithmetic functions are all computable. (Proof left as an exercise!) projection: p(x, y) x constant: c(x) n truncated subtraction: x y { x y if y x 0 if y > x 33 Further examples The following arithmetic functions are all computable. (Proof left as an exercise!) integer division: { integer part of x/y if y > 0 xdivy 0 if y = 0 integer remainder: xmody x y(xdivy) exponentiation base 2: e(x) 2 x logarithm base { 2: greatest y such that 2 log 2 (x) y x if x > 0 0 if x = 0 33

23 Coding programs as numbers 34

24 Turing/Church solution of the Entscheidungsproblem uses the idea that (formal descriptions of) algorithms can be the data on which algorithms act. To realize this idea with Register Machines we have to be able to code RM programs as numbers. (In general, such codings are often called Gödel numberings.) To do that, first we have to code pairs of numbers and lists of numbers as numbers. There are many ways to do that. We fix upon one Numerical coding of pairs For x, y N, define { x, y 2 x (2y + 1) x, y 2 x (2y + 1) 1 So 0b x, y = 0by b x, y = 0by (Notation: 0bx xinbinary.) E.g. 27 = 0b11011 = 0, 13 = 2, 3 36

25 Numerical coding of pairs For x, y N, define { x, y 2 x (2y + 1) x, y 2 x (2y + 1) 1 So 0b x, y = 0by b x, y = 0by 0 1 1, gives a bijection (one-one correspondence) between N N and N., gives a bijection between N N and {n N n = 0}. 36 Numerical coding of lists list N set of all finite lists of natural numbers, using ML notation for lists: empty list: [] list-cons: x :: l list N (given x N and l list N) [x 1, x 2,...,x n ] x 1 :: (x 2 :: ( x n :: [] )) 37

26 Numerical coding of lists list N set of all finite lists of natural numbers, using ML notation for lists. For l list N, define l N by induction on the length { of the list l: [] 0 x :: l x, l = 2 x (2 l + 1) 0b [x 1, x 2,...,x n ] = Hence l l gives a bijection from list N to N. 37 Numerical coding of lists list N set of all finite lists of natural numbers, using ML notation for lists. For l list N, define l N by induction on the length { of the list l: [] 0 x :: l x, l = 2 x (2 l + 1) For example: [3] = 3::[] = 3, 0 = 2 3 ( ) =8 = 0b1000 [1, 3] = 1, [3] = 1, 8 = 34 = 0b [2, 1, 3] = 2, [1, 3] = 2, 34 = 276 = 0b

27 Numerical coding of programs If P is the RM program then its numerical code is L 0 : body 0 L 1 : body 1. L n : body n P [ body 0,..., body n ] where the numerical code body of an instruction body is R + i L j 2i, j defined by: R i L j, L k 2i + 1, j, k HALT 0 38 Any x N decodes to a unique instruction body(x): if x = 0 then body(x) is HALT, else (x > 0 and) let x = y, z in if y = 2i is even, then body(x) is R + i L z, else y = 2i + 1 is odd, let z = j, k in body(x) is R i L j, L k So any e N decodes to a unique program prog(e), called the register machine program with index e: prog(e) L 0 : body(x 0 ). L n : body(x n ) where e = [x 0,...,x n ] 39

28 Example of prog(e) = = 0b }{{} 18 0 s = [18, 0] 18 = 0b10010 = 1, 4 = 1, 0, 2 = R 0 L 0, L 2 0 = HALT So prog(786432) = L 0 : R 0 L 0, L 2 L 1 : HALT N.B. In case e = 0 we have 0 = [], soprog(0) is the program with an empty list of instructions, which by convention we regard as a RM that does nothing (i.e. that halts immediately). 40

29 Universal register machine, U 41

30 High-level specification Universal RM U carries out the following computation, starting with R 0 = 0, R 1 = e (code of a program), R 2 = a (code of a list of arguments) and all other registers zeroed: decode e as a RM program P decode a as a list of register values a 1,...,a n carry out the computation of the RM program P starting with R 0 = 0, R 1 = a 1,...,R n = a n (and any other registers occurring in P set to 0). 42 Mnemonics for the registers of U and the role they play in its program: R 1 P code of the RM to be simulated R 2 A code of current register contents of simulated RM R 3 PC program counter number of the current instruction (counting from 0) R 4 N code of the current instruction body R 5 C type of the current instruction body R 6 R current value of the register to be incremented or decremented by current instruction (if not HALT) R 7 S, R 8 T and R 9 Z are auxiliary registers. R 0 result of the simulated RM computation (if any). 43

31 Overall structure of U s program 1 copy PCth item of list in P to N (halting if PC >length of list); goto 2 2 if N = 0 then halt, else decode N as y, z ; C ::= y; N ::= z; goto 3 {at this point either C = 2i is even and current instruction is R + i L z, or C = 2i + 1 is odd and current instruction is R i L j, L k where z = j, k } 3 copy ith item of list in A to R; goto 4 4 execute current instruction on R; update PC to next label; restore register values to A; goto 1 To implement this, we need RMs for manipulating (codes of) lists of numbers The program START S ::= R HALT to copy the contents of R to S can be implemented by START S R Z HALT Z + R + S + precondition: R = x S = y Z = 0 postcondition: R = x S = x Z = 0 45

32 The program START push X to L HALT to carry out the assignment (X, L) ::=(0, X :: L) can be implemented by START Z + L Z X HALT Z + L + precondition: X = x L = l Z = 0 postcondition: X = 0 L = x, l = 2 x (2l + 1) Z = 0 46 The program START pop L to X HALT EXIT specified by if L = 0 then (X ::= 0; goto EXIT) else let L = x, l in (X ::= x; L ::= l; goto HALT) can be implemented by START X + HALT X L L + L Z Z EXIT Z + L + 47

33 The program for U START T ::= P pop T to N HALT PC pop N to C pop S to R push R to A PC ::= N R + C pop A to R R pop N N + C push R to PC to S 49

34 The halting problem 50

35 Definition. A register machine H decides the Halting Problem if for all e, a 1,...,a n N, startingh with R 0 = 0 R 1 = e R 2 = [a 1,...,a n ] and all other registers zeroed, the computation of H always halts with R 0 containing 0 or 1; moreoverwhenthe computation halts, R 0 = 1 if and only if the register machine program with index e eventually halts when started with R 0 = 0, R 1 = a 1,...,R n = a n and all other registers zeroed. Theorem. No such register machine H can exist. 51 Proof of the theorem Assume we have a RM H that decides the Halting Problem and derive a contradiction, as follows: Let H be obtained from H by replacing START by START Z ::= R 1 push Z to R 2 (where Z is a register not mentioned in H s program). Let C be obtained from H by replacing each HALT (& each erroneous halt) by R0 R 0 +. HALT Let c N be the index of C s program. 52

36 Proof of the theorem Assume we have a RM H that decides the Halting Problem and derive a contradiction, as follows: C started with R 1 = c eventually halts if & only if H started with R 1 = c halts with R 0 = 0 if & only if H started with R 1 = c, R 2 = [c] halts with R 0 = 0 if & only if prog(c) started with R 1 = c does not halt if & only if C started with R 1 = c does not halt contradiction! 52 Computable functions Recall: Definition. f N n N is (register machine) computable if there is a register machine M with at least n + 1 registers R 0, R 1,...,R n (and maybe more) such that for all (x 1,...,x n ) N n and all y N, the computation of M starting with R 0 = 0, R 1 = x 1,...,R n = x n and all other registers set to 0, haltswithr 0 = y if and only if f (x 1,...,x n )=y. Note that the same RM M could be used to compute a unary function (n = 1), or a binary function (n = 2), etc. From now on we will concentrate on the unary case... 53

37 Enumerating computable functions For each e N, letϕ e N N be the unary partial function computed by the RM with program prog(e). So for all x, y N: ϕ e (x) =y holds iff the computation of prog(e) started with R 0 = 0, R 1 = x and all other registers zeroed eventually halts with R 0 = y. Thus e ϕ e defines an onto function from N to the collection of all computable partial functions from N to N. 54 An uncomputable function Let f N N be the partial function with graph {(x,0) ϕ x (x) }. { 0 if ϕ x (x) Thus f (x) = undefined if ϕ x (x) f is not computable, because if it were, then f = ϕ e for some e N and hence if ϕ e (e), then f(e) =0 (by def. of f); so ϕ e (e) =0 (since f = ϕ e ), hence ϕ e (e) if ϕ e (e), then f(e) (since f = ϕ e ); so ϕ e (e) (by def. of f) contradiction! So f cannot be computable. 55

38 (Un)decidable sets of numbers Given a subset S N, itscharacteristic { function 1 if x S χ S N N is given by: χ S (x) 0 if x / S. 56 (Un)decidable sets of numbers Definition. S N is called (register machine) decidable if its characteristic function χ S N N is a register machine computable function. Otherwise it is called undecidable. So S is decidable iff there is a RM M with the property: for all x N, M started with R 0 = 0, R 1 = x and all other registers zeroed eventually halts with R 0 containing 1 or 0; andr 0 = 1 on halting iff x S. Basic strategy: to prove S N undecidable, try to show that decidability of S would imply decidability of the Halting Problem. For example... 56

39 Claim: S 0 {e ϕ e (0) } is undecidable. Proof (sketch): Suppose M 0 is a RM computing χ S0.FromM 0 s program (using the same techniques as for constructing a universal RM) we can construct a RM H to carry out: let e = R 1 and [a 1,...,a n ] = R 2 in R 1 ::= (R 1 ::= a 1 ) ; ; (R n ::= a n ) ; prog(e) ; R 2 ::= 0; run M 0 Then by assumption on M 0, H decides the Halting Problem contradiction. So no such M 0 exists, i.e. χ S0 is uncomputable, i.e. S 0 is undecidable. 57 Claim: S 1 {e ϕ e atotalfunction} is undecidable. Proof (sketch): Suppose M 1 is a RM computing χ S1.FromM 1 s program we can construct a RM M 0 to carry out: let e = R 1 in R 1 ::= R 1 ::= 0;prog(e) ; run M 1 Then by assumption on M 1, M 0 decides membership of S 0 from previous example (i.e. computes χ S0 ) contradiction. So no such M 1 exists, i.e. χ S1 is uncomputable, i.e. S 1 is undecidable. 58

40 Turing machines 59

41 Algorithms, informally No precise definition of algorithm at the time Hilbert posed the Entscheidungsproblem, justexamples. Common features of the examples: finite description of the procedure in terms of elementary operations deterministic (next step uniquely determined if there is one) procedure may not terminate on some input data, but we can recognize when it does terminate and what the result is. e.g. multiply two decimal digits by looking up their product in a table 60 Register Machine computation abstracts away from any particular, concrete representation of numbers (e.g. as bit strings) and the associated elementary operations of increment/decrement/zero-test. Turing s original model of computation (now called a Turing machine)is more concrete:even numbers have to be represented in terms of a fixed finite alphabet of symbols and increment/decrement/zero-test programmed in terms of more elementary symbol-manipulating operations. 61

42 Turing machines, informally machine is in one of afinitesetofstates q tape symbol being scanned by tape head special left endmarker symbol special blank symbol linear tape, unbounded to right, divided into cells containing a symbol from a finite alphabet of tape symbols. Only finitely many cells contain non-blank symbols. 62 Turing machines, informally q Machine starts with tape head pointing to the special left endmarker. Machine computes in discrete steps, each of which depends only on current state (q) andsymbolbeingscannedbytapehead(0). Action at each step is to overwrite the current tape cell with a symbol, move left or right one cell, or stay stationary, and change state. 62

43 Turing Machines are specified by: Q, finitesetofmachinestates Σ, finite set of tape symbols (disjoint from Q) containing distinguished symbols (left endmarker) and (blank) s Q, aninitial state δ (Q Σ) (Q {acc, rej}) Σ {L, R, S}, a transition function, satisfying: for all q Q, thereexistsq Q {acc, rej} with δ(q, ) =(q,, R) (i.e. left endmarker is never overwritten and machine always moves to the right when scanning it) 63 Example Turing Machine M =(Q, Σ, s, δ) where states Q = {s, q, q } (s initial) symbols Σ = {,,0,1} transition function δ (Q Σ) (Q {acc, rej}) Σ {L, R, S}: δ 0 1 s (s,, R) (q,, R) (rej, 0, S) (rej, 1, S) q (rej,, R) (q,0,l) (q,1,r) (q,1,r) q (rej,, R) (acc,, S) (rej,0,s) (q,1,l) 64

44 Turing machine computation Turing machine configuration: (q, w, u) where q Q {acc, rej} =currentstate w =non-emptystring(w = va) of tape symbols under and to the left of tape head, whose last element (a) iscontentsofcell under tape head u =(possiblyempty)stringoftapesymbolstotherightoftape head (up to some point beyond which all symbols are ) Initial configurations: (s,, u) 65 Turing machine computation Given a TM M =(Q, Σ, s, δ), wewrite (q, w, u) M (q, w, u ) to mean q = acc, rej, w = va (for some v, a) and either δ(q, a) =(q, a, L), w = v, andu = a u or δ(q, a) =(q, a, S), w = va and u = u or δ(q, a) =(q, a, R), u = a u is non-empty, w = va a and u = u or δ(q, a) =(q, a, R), u = ε is empty, w = va and u = ε. 65

45 Turing machine computation A computation of a TM M is a (finite or infinite) sequence of configurations c 0, c 1, c 2,... where c 0 =(s,, u) is an initial configuration c i M c i+1 holds for each i = 0, 1,.... The computation does not halt if the sequence is infinite halts if the sequence is finite and its last element is of the form (acc, w, u) or (rej, w, u). 65 Example Turing Machine M =(Q, Σ, s, δ) where states Q = {s, q, q } (s initial) symbols Σ = {,,0,1} transition function δ (Q Σ) (Q {acc, rej}) Σ {L, R, S}: δ 0 1 s (s,, R) (q,, R) (rej, 0, S) (rej, 1, S) q (rej,, R) (q,0,l) (q,1,r) (q,1,r) q (rej,, R) (acc,, S) (rej,0,s) (q,1,l) Claim: the computation of M starting from configuration (s,, 1 n 0) halts in configuration (acc,,1 n+1 0). 66

46 The computation of M starting from configuration (s,, 1 n 0): (s,, 1 n 0) M (s,,1 n 0) M. (q, 1, 1 n 1 0) M (q, 1 n,0) M (q, 1 n 0, ε) M (q, 1 n+1, ε) M. (q, 1 n+1,0) M (q,,1 n+1 0) M (acc,,1 n+1 0) 67 Theorem. The computation of a Turing machine M can be implemented by a register machine. Proof (sketch). Step 1: fix a numerical encoding of M s states, tape symbols, tape contents and configurations. Step 2: implement M s transition function (finite table) using RM instructions on codes. Step 3: implement a RM program to repeatedly carry out M. 68

47 Step 1 Identify states and tape symbols with particular numbers: acc = 0 = 0 rej = 1 = 1 Q = {2, 3,..., n} Σ = {0, 1,..., m} Code configurations c =(q, w, u) by: c = [q, [a n,...,a 1 ], [b 1,...,b m ] ] where w = a 1 a n (n > 0) andu = b 1 b m (m 0) say. 69 Step 2 Using registers Q =currentstate A =currenttapesymbol D =currentdirectionoftapehead (with L = 0, R = 1 and S = 2, say) one can turn the finite table of (argument,result)-pairs specifying δ into a RM program (Q, A, D) ::= δ(q, A) so that starting the program with Q = q, A = a, D = d (and all other registers zeroed), it halts with Q = q, A = a, D = d,where(q, a, d )=δ(q, a). 70

48 Step 3 The next slide specifies a RM to carry out M s computation. It uses registers C =codeofcurrentconfiguration W =codeoftapesymbolsatandleftoftapehead(reading right-to-left) U =codeoftapesymbolsrightoftapehead(reading left-to-right) Starting with C containing the code of an initial configuration (and all other registers zeroed), the RM program halts if and only if M halts; and in that case C holds the code of the final configuration. 71 START HALT yes [Q,W,U] ::=C Q<2? no pop W to A (Q,A,D)::=δ(Q,A) C::= [Q,W,U] push A to U D push B to W pop U D push A to B to W 72

49 Computable functions Recall: Definition. f N n N is (register machine) computable if there is a register machine M with at least n + 1 registers R 0, R 1,...,R n (and maybe more) such that for all (x 1,...,x n ) N n and all y N, the computation of M starting with R 0 = 0, R 1 = x 1,...,R n = x n and all other registers set to 0, haltswithr 0 = y if and only if f (x 1,...,x n )=y. 73 We ve seen that a Turing machine s computation can be implemented by a register machine. The converse holds: the computation of a register machine can be implemented by a Turing machine. To make sense of this, we first have to fix a tape representation of RM configurations and hence of numbers and lists of numbers... 74

50 Tape encoding of lists of numbers Definition. AtapeoverΣ = {,,0,1} codes a list of numbers if precisely two cells contain 0 and the only cells containing 1 occur between these. Such tapes look like: 01 1 }{{} 1 } 1 {{} 1} 1 {{} 0 }{{ } n 1 n 2 n k which corresponds to the list [n 1, n 2,...,n k ]. all s 75 Turing computable function Definition. f N n N is Turing computable if and only if there is a Turing machine M with the following property: Starting M from its initial state with tape head on the left endmarker of a tape coding [0, x 1,...,x n ], M halts if and only if f (x 1,...,x n ), andinthatcasethefinaltape codes a list (of length 1) whosefirstelementis y where f (x 1,...,x n )=y. 76

51 Theorem. ApartialfunctionisTuringcomputableifand only if it is register machine computable. Proof (sketch). We ve seen how to implement any TM by a RM. Hence f TM computable implies f RM computable. For the converse, one has to implement the computation of a RM in terms of a TM operating on a tape coding RM configurations. To do this, one has to show how to carry out the action of each type of RM instruction on the tape. It should be reasonably clear that this is possible in principle, even if the details (omitted) are tedious. 77 Notions of computability Church (1936): λ-calculus [see later] Turing (1936): Turing machines. Turing showed that the two very different approaches determine the same class of computable functions. Hence: Church-Turing Thesis. Every algorithm [in intuitive sense of Lect. 1] can be realized as a Turing machine. 78

52 Notions of computability Church-Turing Thesis. Every algorithm [in intuitive sense of Lect. 1] can be realized as a Turing machine. Further evidence for the thesis: Gödel and Kleene (1936): partial recursive functions Post (1943) and Markov (1951): canonical systems for generating the theorems of a formal system Lambek (1961) and Minsky (1961): register machines Variations on all of the above (e.g. multiple tapes, non-determinism, parallel execution... ) All have turned out to determine the same collection of computable functions. 78 Notions of computability Church-Turing Thesis. Every algorithm [in intuitive sense of Lect. 1] can be realized as a Turing machine. In rest of the course we ll look at Gödel and Kleene (1936): partial recursive functions ( branch of mathematics called recursion theory) Church (1936): λ-calculus ( branch of CS called functional programming) 78

53 Aim Amoreabstract,machine-independentdescriptionofthe collection of computable partial functions than provided by register/turing machines: they form the smallest collection of partial functions containing some basic functions and closed under some fundamental operations for forming new functions from old composition, primitive recursion and minimization. The characterization is due to Kleene (1936), building on work of Gödel and Herbrand. 79 Basic functions Projection functions, proj n i N n N: proj n i (x 1,...,x n ) x i Constant functions with value 0, zero n N n N: zero n (x 1,...,x n ) 0 Successor function, succ N N: succ(x) x

54 Basic functions are all RM computable: Projection proji n is computed by START R 0 ::= R i HALT Constant zero n is computed by START HALT Successor succ is computed by START R + 1 R 0 ::= R 1 HALT 81 Composition Composition of f N n N with g 1,...,g n N m N is the partial function f [g 1,...,g n ] N m N satisfying for all x 1,...,x m N f [g 1,...,g n ](x 1,...,x m ) f (g 1 (x 1,...,x m ),...,g n (x 1,...,x m )) where is Kleene equivalence of possibly-undefined expressions: LHS RHS means either both LHS and RHS are undefined, or they are both defined and are equal. 82

55 Composition Composition of f N n N with g 1,...,g n N m N is the partial function f [g 1,...,g n ] N m N satisfying for all x 1,...,x m N f [g 1,...,g n ](x 1,...,x m ) f (g 1 (x 1,...,x m ),...,g n (x 1,...,x m )) So f [g 1,...,g n ](x 1,...,x m )=z iff there exist y 1,...,y n with g i (x 1,...,x m )=y i (for i = 1..n) and f (y 1,...,y n )=z. N.B. in case n = 1, wewrite f g 1 for f [g 1 ]. 82 Composition f [g 1,...,g n ] is computable if f and g 1,...,g n are. { { F f(y1,...,y Proof. Given RM programs computing n ) Gi g i (x 1,...,x m ) in R 0 { { R1,...,R starting with n y1,...,y set to n,thenthenextslide R 1,...,R m x 1,...,x m specifies a RM program computing f [g 1,...,g n ](x 1,...,x m ) in R 0 starting with R 1,...,R m set to x 1,...,x m. (Hygiene [caused by the lack of local names for registers in the RM model of computation]: we assume the programs F, G 1,...,G n only mention registers up to R N (where N max{n, m}) andthat X 1,...,X m, Y 1,...,Y n are some registers R i with i > N.) 82

56 START (X 1,...,X m )::=(R 1,...,R m ) G 1 Y 1 ::= R 0 (R 0,...,R N )::=(0,...,0) (R 1,...,R m )::=(X 1,...,X m ) G 2 Y 2 ::= R 0 (R 0,...,R N )::=(0,...,0) (R 1,...,R m )::=(X 1,...,X m ) G n Y n ::= R 0 (R 0,...,R N )::=(0,...,0) (R 1,...,R n )::=(Y 1,...,Y n ) F 83

57 Partial recursive functions 84

58 Examples of recursive definitions { f 1 (0) 0 f 1 (x + 1) f 1 (x)+(x + 1) f 2 (0) 0 f 2 (1) 1 f 2 (x + 2) f 2 (x)+f 2 (x + 1) { f 3 (0) 0 f 3 (x + 1) f 3 (x + 2)+1 f 1 (x) = sum of 0, 1, 2,..., x f 2 (x) =xth Fibonacci number f 3 (x) undefined except when x = 0 f 4 (x) if x > 100 then x 10 else f 4 ( f 4 (x + 11)) f 4 is McCarthy s "91 function", which maps x to 91 if x 100 and to x 10 otherwise 85 Primitive recursion Theorem. Given f N n N and g N n+2 N, there is a unique h N n+1 N satisfying { h( x,0) f ( x) h( x, x + 1) g( x, x, h( x, x)) for all x N n and x N. We write ρ n ( f, g) for h and call it the partial function defined by primitive recursion from f and g. 87

59 Primitive recursion Theorem. Given f N n N and g N n+2 N, there is a unique h N n+1 N satisfying { h( x,0) f ( x) ( ) h( x, x + 1) g( x, x, h( x, x)) for all x N n and x N. Proof (sketch). Existence: theset h {( x, x, y) N n+2 y 0, y 1,...,y x f( x) =y 0 ( x 1 i=0 g( x, i, y i)=y i+1 ) y x = y} defines a partial function satisfying ( ). Uniqueness: ifh and h both satisfy ( ), thenonecanproveby induction on x that x (h( x, x) =h ( x, x)). 88 Example: addition Addition add N 2 N satisfies: { add(x 1,0) x 1 add(x 1, x + 1) add(x 1, x)+1 So add = ρ 1 ( f, g) where { f (x 1 ) x 1 g(x 1, x 2, x 3 ) x Note that f = proj 1 1 and g = succ proj3 3 ;soadd can be built up from basic functions using composition and primitive recursion: add = ρ 1 (proj 1 1, succ proj3 3). 89

60 Example: predecessor Predecessor pred N N satisfies: { pred(0) 0 pred(x + 1) x So pred = ρ 0 ( f, g) where { f () 0 g(x 1, x 2 ) x 1 Thus pred can be built up from basic functions using primitive recursion: pred = ρ 0 (zero 0, proj 2 1). 90 Example: multiplication Multiplication mult N 2 N satisfies: { mult(x 1,0) 0 mult(x 1, x + 1) mult(x 1, x)+x 1 and thus mult = ρ 1 (zero 1, add (proj 3 3, proj3 1 )). So mult can be built up from basic functions using composition and primitive recursion (since add can be). 91

61 Definition. A[partial]function f is primitive recursive ( f PRIM) ifitcanbebuiltupinfinitelymanysteps from the basic functions by use of the operations of composition and primitive recursion. In other words, the set PRIM of primitive recursive functions is the smallest set (with respect to subset inclusion) of partial functions containing the basic functions and closed under the operations of composition and primitive recursion. 92 Definition. A [partial] function f is primitive recursive ( f PRIM) ifitcanbebuiltupinfinitelymanysteps from the basic functions by use of the operations of composition and primitive recursion. Every f PRIM is a total function, because: all the basic functions are total if f, g 1,...,g n are total, then so is f (g 1,...,g n ) [why?] if f and g are total, then so is ρ n ( f, g) [why?] 92

62 Definition. A[partial]function f is primitive recursive ( f PRIM) ifitcanbebuiltupinfinitelymanysteps from the basic functions by use of the operations of composition and primitive recursion. Theorem. Every f PRIM is computable. Proof. Already proved: basic functions are computable; composition preserves computability. So just have to show: ρ n ( f, g) N n+1 N computable if f N n N and g N n+2 N are. Suppose f and g are computed by RM programs F and G (with our usual I/O conventions). Then the RM specified on the next slide computes ρ n ( f, g). (WeassumeX 1,...,X n+1, C are some registers not mentioned in F and G; andthatthelatteronlyuseregisters R 0,...,R N,whereN n + 2.) 92 START (X 1,...,X n+1,r n+1 )::=(R 1,...,R n+1,0) F C + C=X n+1? yes no HALT (R 1,...,R n,r n+1,r n+2 )::=(X 1,...,X n,c,r 0 ) G (R 0,R n+3,...,r N )::=(0,0,...,0) 93

63 Aim Amoreabstract,machine-independentdescriptionofthe collection of computable partial functions than provided by register/turing machines: they form the smallest collection of partial functions containing some basic functions and closed under some fundamental operations for forming new functions from old composition, primitive recursion and minimization. The characterization is due to Kleene (1936), building on work of Gödel and Herbrand. 94 Minimization Given a partial function f N n+1 N, define µ n f N n N by µ n f ( x) least x such that f ( x, x) =0 and for each i = 0,..., x 1, f ( x, i) is defined and > 0 (undefined if there is no such x) In other words µ n f = {( x, x) N n+1 y 0,...,y x x x 1 ( f( x, i) =y i ) ( y i > 0) y x = 0} i=0 i=0 95

64 Example of minimization integer part of x 1 /x 2 least x 3 such that (undefined if x 2 =0) x 1 < x 2 (x 3 + 1) where f N 3 N is f (x 1, x 2, x 3 ) µ 2 f (x 1, x 2 ) { 1 if x 1 x 2 (x 3 + 1) 0 if x 1 < x 2 (x 3 + 1) (In fact, if we make the integer part of x 1 /x 2 functiontotalby defining it to be 0 when x 2 = 0, itcanbeshowntobeinprim.) 96 Definition. A partial function f is partial recursive ( f PR) ifitcanbebuiltupinfinitelymanystepsfrom the basic functions by use of the operations of composition, primitive recursion and minimization. In other words, the set PR of partial recursive functions is the smallest set (with respect to subset inclusion) of partial functions containing the basic functions and closed under the operations of composition, primitive recursion and minimization. 97

65 Definition. A partial function f is partial recursive ( f PR) ifitcanbebuiltupinfinitelymanystepsfrom the basic functions by use of the operations of composition, primitive recursion and minimization. Theorem. Every f PR is computable. Proof. Just have to show: µ n f N n N is computable if f N n+1 N is. Suppose f is computed by RM program F (with our usual I/O conventions). Then the RM specified on the next slide computes µ n f. (We assume X 1,...,X n, C are some registers not mentioned in F; and that the latter only uses registers R 0,...,R N,whereN n + 1.) 97 START (X 1,...,X n )::=(R 1,...,R n ) (R 1,...,R n,r n+1 )::=(X 1,...,X n,c) C + (R 0,R n+2,...,r N )::=(0,0,...,0) F R 0 R 0 ::=C HALT 98

66 Computable = partial recursive Theorem. Not only is every f PR computable, but conversely, every computable partial function is partial recursive. Proof (sketch). Let f be computed by RM M. Recallhowwecoded instantaneous configurations c =(l, r 0,...,r n ) of M as numbers [l, r 0,...,r n ]. Itispossibletoconstructprimitiverecursivefunctions lab, val 0, next M N N satisfying lab( [l, r 0,...,r n ] ) =l val 0 ( [l, r 0,...,r n ] ) =r 0 next M ( [l, r 0,...,r n ] ) =code of M s next configuration (Showing that next M PRIM is tricky proof omitted.) L8 99 Proof sketch, cont. Writing x for x 1,...,x n,letconfig M ( x, t) be the code of M s configuration after t steps, starting with initial register values R 0 = 0, R 1 = x 1,...,R n = x n.it sinprim because: { config M ( x,0) = [0, 0, x] config M ( x, t + 1) = next M (config M ( x, t)) Can assume M has a single HALT as last instruction, Ith say (and no erroneous halts). Let halt M ( x) be the number of steps M takes to halt when started with initial register values x (undefined if M does not halt). It satisfies halt M ( x) least t such that I lab(config M ( x, t)) = 0 and hence is in PR (because lab, config M, I () PRIM). L8 100

67 Definition. A partial function f is partial recursive ( f PR) ifitcanbebuiltupinfinitelymanystepsfrom the basic functions by use of the operations of composition, primitive recursion and minimization. The members of PR that are total are called recursive functions. Fact: there are recursive functions that are not primitive recursive. For example Ackermann s function There is a (unique) function ack N 2 N satisfying ack(0, x 2 ) = x ack(x 1 + 1, 0) = ack(x 1,1) ack(x 1 + 1, x 2 + 1) = ack(x 1, ack(x 1 + 1, x 2 )) ack is computable, hence recursive [proof: exercise]. Fact: ack grows faster than any primitive recursive function f N 2 N: N f x 1, x 2 > N f ( f (x 1, x 2 ) < ack(x 1, x 2 )). Hence ack is not primitive recursive. 102

68 Lambda calculus 103

69 Notions of computability Church (1936): λ-calculus Turing (1936): Turing machines. Turing showed that the two very different approaches determine the same class of computable functions. Hence: Church-Turing Thesis. Every algorithm [in intuitive sense of Lect. 1] can be realized as a Turing machine. 104 λ-terms, M are built up from a given, countable collection of variables x, y, z,... by two operations for forming λ-terms: λ-abstraction: (λx.m) (where x is a variable and M is a λ-term) application: (M M ) (where M and M are λ-terms). Some random examples of λ-terms: x (λx.x) ((λy.(xy))x) (λy.((λy.(xy))x)) 105

70 Free and bound variables In λx.m, we call x the bound variable and M the body of the λ-abstraction. An occurrence of x in a λ-term M is called binding if in between λ and. (e.g. (λx.yx) x) bound if in the body of a binding occurrence of x (e.g. (λx.y x) x) free if neither binding nor bound (e.g. (λx.yx)x). 106 Free and bound variables Sets of free and bound variables: FV(x) = {x} FV(λx.M) = FV(M) {x} FV(M N) = FV(M) FV(N) BV(x) = BV(λx.M) = BV(M) {x} BV(M N) = BV(M) BV(N) If FV(M) =, M is called a closed term, or combinator. 106

71 α-equivalence M = α M λx.m is intended to represent the function f such that f (x) =M for all x. So the name of the bound variable is immaterial: if M = M{x /x} is the result of taking M and changing all occurrences of x to some variable x # M, thenλx.m and λx.m both represent the same function. For example, λx.x and λy.y represent the same function (the identity function). 107 α-equivalence M = α M is the binary relation inductively generated by the rules: x = α x z # (M N) M{z/x} = α N{z/y} λx.m = α λy.n M = α M N = α N MN= α M N where M{z/x} is M with all occurrences of x replaced by z. 107

72 For example: α-equivalence M = α M λx.(λxx.x) x = α λy.(λxx.x)x because (λzx.z)x = α (λxx.x)x because λzx.z = α λxx.x and x = α x because λx.u = α λx.u and x = α x because u = α u and x = α x. 107 α-equivalence M = α M Fact: = α is an equivalence relation (reflexive, symmetric and transitive). We do not care about the particular names of bound variables, just about the distinctions between them. So α-equivalence classes of λ-terms are more important than λ-terms themselves. Textbooks (and these lectures) suppress any notation for α-equivalence classes and refer to an equivalence class via a representative λ-term (look for phrases like we identify terms up to α-equivalence or we work up to α-equivalence ). For implementations and computer-assisted reasoning, there are various devices for picking canonical representatives of α-equivalence classes (e.g. de Bruijn indexes, graphical representations,... ). 107

73 Substitution N[M/x] x[m/x] = M y[m/x] = y if y = x (λy.n)[m/x] = λy.n[m/x] if y # (Mx) (N 1 N 2 )[M/x] = N 1 [M/x] N 2 [M/x] Side-condition y # (M x) (y does not occur in M and y = x) makessubstitution capture-avoiding. E.g. if x = y = z = x (λy.x)[y/x] = α (λz.x)[y/x] =λz.y In fact N N[M/x] induces a totally defined function from the set of α-equivalence classes of λ-terms to itself. 108 β-reduction Recall that λx.m is intended to represent the function f such that f (x) =M for all x. We can regard λx.m as a function on λ-terms via substitution: map each N to M[N/x]. So the natural notion of computation for λ-terms is given by stepping from a β-redex (λx.m)n to the corresponding β-reduct M[N/x] 109

74 β-reduction One-step β-reduction, M M : (λx.m)n M[N/x] M M MN M N M M λx.m λx.m M M NM NM N = α M M M M = α N N N 109 E.g. β-reduction ((λy.λz.z)u)y (λx.xy)((λy.λz.z)u) (λz.z)y y (λx.xy)(λz.z) E.g. of up to α-equivalence aspect of reduction: (λx.λy.x)y = α (λx.λz.x)y λz.y 109

75 Many-step β-reduction, M M : M = α M M M (no steps) M M M M (1 step) M M M M M M (1 more step) E.g. (λx.xy)((λyz.z)u) y (λx.λy.x)y λz.y 110 β-conversion M = β N Informally: M = β N holds if N can be obtained from M by performing zero or more steps of α-equivalence, β-reduction, or β-expansion (= inverse of a reduction). E.g. u ((λxy. vx)y) = β (λx. ux)(λx. vy) because (λx. ux)(λx. vy) u(λx. vy) and so we have u ((λxy. vx)y) = α u ((λxy. vx)y) u(λy. vy) reduction = α u(λx. vy) (λx. ux)(λx. vy) expansion 111

76 β-conversion M = β N is the binary relation inductively generated by the rules: M = α M M M M = β M M = β M M = β M M = β M M = β M M = β M M = β M M = β M λx.m = β λx.m M = β M N = β N MN= β M N 111 Church-Rosser Theorem Theorem. is confluent, thatis,ifm 1 M M 2, then there exists M such that M 1 M M 2. Corollary. M 1 = β M 2 iff M (M 1 M M 2 ). Proof. = β satisfies the rules generating ; som M implies M = β M.ThusifM 1 M M 2,thenM 1 = β M = β M 2 and so M 1 = β M 2. Conversely, the relation {(M 1, M 2 ) M (M 1 M M 2 )} satisfies the rules generating = β :theonlydifficult case is closure of the relation under transitivity and for this we use the Church-Rosser theorem. Hence M 1 = β M 2 implies M (M 1 M M 2 ). 112

77 β-normal Forms Definition. A λ-term N is in β-normal form (nf) ifit contains no β-redexes (no sub-terms of the form (λx.m)m ). M has β-nf N if M = β N with N a β-nf. Note that if N is a β-nf and N N,thenitmustbethatN = α N (why?). Hence if N 1 = β N 2 with N 1 and N 2 both β-nfs, then N 1 = α N 2.(For if N 1 = β N 2,thenbyChurch-RosserN 1 M N 2 for some M, so N 1 = α M = α N 2.) So the β-nf of M is unique up to α-equivalence if it exists. 113 Non-termination Some λ terms have no β-nf. E.g. Ω (λx.xx)(λx.xx) satisfies Ω (xx)[(λx.xx)/x] =Ω, Ω M implies Ω = α M. So there is no β-nf N such that Ω = β N. Atermcanpossessbothaβ-nf and infinite chains of reduction from it. E.g. (λx.y)ω y, butalso(λx.y)ω (λx.y)ω. 114