Combinators & Lambda Calculus

Transcription

1 Combinators & Lambda Calculus

2 Abstracting 1/16 three apples plus two pears = five fruits concrete 3+2 = 5 abstract objects a+b = b+a a (b c) = (a b) c abstract quantities abstract operations a, b[r(a, b) R(b, a)] a[ R(a, a)] abstract relation Sxyz = xz(yz) abstract meaning

3 Words, language, theory 2/16 Remember that for a language L over an alphabet Σ one has L Σ Σ consists of all strings, possibly nonsensical L Σ chooses in some way meaningful strings called sentences often such a language is given by a grammar With a theory T we go one step further: A theory in a language L is just a subset T L selecting a set of correct sentences often such a theory is given by an axiomatic system

4 Words, language, theory 3/16 3)( 2 + meaningless = 7 2 meaningful, incorrect = 5 2 meaningful, correct T L Σ Σ Σ L T

5 Combinators 4/16 Σ CL = {I,K,S,x,,),(,=} We introduce several simple regular grammars over Σ CL. (i) constant := I K S (ii) variable := x variable (iii) term := constant variable (term term) (iv) formula := term = term Intuition: in (FA) the term F stands for a function and A for an argument

6 Combinatory Logic (Schönfinkel 1920 [1924]) 5/16 Axioms IP = P (I) KPQ = P (K) SPQR = PR(QR) (S) Deduction rules P = P P = Q Q = P P = Q,Q = R P = R P = Q PR = QR P = Q RP = RQ Here P,Q,R denote arbitrary terms IP stands for (IP), KPQ for ((KP)Q) and SPQR for (((SP)Q)R) In general PQ 1...Q n (..((PQ 1 )Q 2 )...Q n ) (association to the left)

7 Some magic with combinators 6/16 Proposition. (i) Let D SII. Then (doubling) Dx = CL xx. (ii) Let B S(KS)K. Then (composition) Bfgx = CL f(gx). (iii) Let L D(BDD). Then (self-doubling, life!) L = CL LL. Proof. (i) Dx SIIx = Ix(Ix) = xx. (ii) Bf gx S(KS)Kf gx = KSf(Kf)gx = S(Kf)gx = Kf x(gx) = f(gx). We want to understand and preferably also to control this! (iii) L D(BDD) = BDD(BDD) = D(D(BDD)) DL = LL.

8 First insight 7/16 Lemma. For every term P and every variable x, there is a term Q such that x does not occur in Q and Qx = CL P. We denote this term Q constructed in the proof as [x]p. Proof. Induction on the complexity of P. Case 1. P is a constant or a variable. Subcase 1.1 P C with C {I,K,S}. Take [x]c KC. Then indeed Subcase 1.2 P x. Take [x]x I. Then ([x]c)x = CL KCx = CL C. ([x]x)x Ix = CL x. Subcase 1.3 P y x. Take [x]y Kx. Then indeed ([x]y)x Kyx = CL y. Case 2. P UV. Take [x](uv) S([x]U)([x]V). Then indeed ([x](uv))x S([x]U)([x]V)x = CL (([x]u)x)(([x]v)x) = CL UV.

9 Algorithms 8/16 The previous proof gave P [x]p ([x]p)x = P? C KC KCx = C x I Ix = x y x Ky Kyx = y UV S([x]U)([x]V) S([x]U)([x]V)x = (([x]u)x)(([x]v)x) = UV More efficient algorithm P x P with x / P UV [x]p I KP S([x]U)([x]V)

10 Lambda Calculus: intended meaning 9/16 The meaning of λx.3x is the function x 3x that assigns to x the value 3x (3 times x) So according to this intended meaning we have (λx.3x)(6) = 18. The parentheses around the 6 are usually not written: Principal axiom So λx.m intends to capture [x]m (λx.3x)6 = 18 (λx.m)n = β M[x: = N]

11 Language and Theory (Church [1932]) 10/16 Alphabet Σ = {x,,(,),λ,=} Language variable : = x variable Theory term : = variable (term term) (λvariable term) formula : = term = term Axioms (λxm)n = M[x: = N] M = M Rules M = N N = M M = N,N = L M = N M = N ML = NL M = N LM = LN M = N λxm = λxn

12 Bureaucracy 11/16 Substitution M M[x: = N] x N y y PQ (P[x: = N])(Q[x: = N]) λxp λxp λyp λy(p[x: = N]) where y x Association to the left Associating to the right PQ 1...Q n (..((PQ 1 )Q 2 )...Q n ). λx 1...x n.m (λx 1 (λx 2 (..(λx n (M))..)))). Outer parentheses are often omitted. For example (λx.x)y ((λxx)y)

13 Examples 12/16 I (λx.x) IX = β X K (λxy.x) KXY = β X S (λxyz.xz(yz)) SXYZ = β XZ(YZ) D (λx.xx) DX = β XX B (λxyz.x(yz) BXYZ = β X(YZ) Set of lambda terms: Λ

14 Fixed point theorem 13/16 Theorem. For all F Λ there is an M Λ such that FM = β M Proof. Defines W λx.f(xx) and M WW. Then M WW (λx.f(xx))w = F(WW) FM. Corollary. For any context C[ x,m] there exists a M such that M X = C[ X,M]. Proof. M can be taken the fixed point of λm x.c[ x,m]. Then M X = (λm x.c[ x,m])m X = C[ X,M].

15 From Russell Paradox to Fixed point Theorem 14/16 The axiom of (unlimited) comprehension (essentially in Frege [1993]) is A = {x P(x)} exists, i.e. a.[a A P(a)] But then we can derive the Russell Paradox. Define then so R = {x x / x} a.[a R a / a] R R R / R, a contradiction. Haskell Curry found the fixed point theorem analyzing the Russell paradox: Writing x A as Ax (identifying sets and predicates) one obtains a.[ra (aa)] RR (RR) the meaning of R is λx. (xx), as Rx (xx) Now RR is the fixed point of, leading to the proof of the fixed point theorem

16 Consequences 15/16 We can construct terms Y,L,O,P such that Yf = f(yf) L = LL Ox = O Q = QI; Px = xp. producing fixed points; take L YD; take O YK;

17 References 16/16 A Church [1936] An unsolvable problem of elementary number theory, Am. J. Math., 58, HB Curry [1930] Grundlagen der Kombinatorischen Logik, American J. Mathematics, 52 (3), HB Curry [1934] Functionality in combinatory logic. Proc. Nat. Ac. Sc. USA, 20, G Frege [1893, 1903/1966] Grundgesetze der Arithmetik. Hildesheim: Georg Olms Verlag. M Schönfinkel [1924] On the building blocks of mathematical logic. In: van Heijenoort, Ed., 1967, From Frege to Gödel: A Source Book in Mathematical Logic, Cambridge, MA: Harvard University Press,