By V.E. Wolfengagen
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Additional resources for Combinatory Logic in Programming. Computations with Objects Through Examples and Exercises
Derive via K and S the object with combinatory characteristic: Ia = a, (I) using postulates α, β, µ, ν, σ, τ , ξ of λ-conversion. Solution. I–1. b a=b ; ca = cb a = b; b = c ; (σ) a=c a=b . b=a I–2. xz(yz). 36 C HAPTER 1: P RELIMINARIES I–3. Using schemes (K) and (S), get confidence in that: a = Ka(Ka) = SKKa. (K) (S) Checking. Make sure, that actually I = SKK. Let v = empty (empty object). I–1. SKKa = Ka(Ka), because in scheme (S) can be set x = K, y = K, z = a. Then it is evident, that by postulate (α): Sxyz = SKKa, xz(yz) = Ka(Ka), SKKa = Ka(Ka).
Answer. e. I = SKK. 5 Variants of task ‡ Work 1. Derive via K and S the objects with given combinatory characteristics: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) Babc Cabc W ab Ψabcd C  abcd C abcd B 2 abcd Ya C  abcde C abcde B 3 abcde Φabcd = = = = = = = = = = = = a(bc), acb, abb, a(bc)(bd), acdb, adbc, a(bcd), a(Y a) (prove, that Y = W S(BW B)), acdeb, aebcd, a(bcde), a(bd)(cd). ‡ Work 2. f (x(x)). ) 38 C HAPTER 1: P RELIMINARIES ‡ Work 3. f (x(x)). ‡ Work 4. ): 1) Ξ 2) F 3) P 4) & 5) ∨ 6) ¬ 7) ∃∗ = = = = = = = C(BCF )I, B(CB 2 B)Ξ, ΨΞK, B 2 (CΞI)(C(BB 2 P )P ), B 2 (CΞI)(C(B 2 B(B(Φ&))P )P ), CP (ΠI), B(W (B 2 (ΦP )CΞ))K, where ∃[a] = ∃∗ [a], ∃ = ∃[I].
W –3. The additional two variants of derivations for W are as follows: abb = ab(Kba) = ab(CKab) Sa(CKa)b = SS(CK)ab abb = = = = = ab(Kb(Kb)) ab(SKKb) Sa(SKK)b Sa(K(SKK)a)b SS(K(SKK))ab. Answer. The object W with a characteristic W ab = abb is as follows: W = CSI ( = SS(CK) = SS(K(SKK)) ). 4. Derive the expression for combinator Ψ. Task formulation. Derive combinator Ψ with the following characteristic: Ψabcd = a(bc)(bd). (Ψ) Solution. Ψ–1. List the postulates for conversion relation.