Please solve this with clarification No coding is needed Consider the list: (Po, 1.n-1) being represented in
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Please solve this with clarification
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Consider the list: (Po, 1₁.n-1) being represented in the λ-calculus as follows: (Po): λx.((x0)4) (Do, 0₁): λx.((x₂)(₁)) (Po, P₁, P₂): λx.((x0) 0₁, 02)) (Po, P1, Pn-1): Ax.((x0) 0₁, 02, 0.1)) is a "dummy" 'end of list' delimiter. It can be any λ-expression. The idea, in Exercise 7.1 on the page before, to let T and F select into a list of length 2 and yield 1st, respectively the second element, is now iterated: T:Ax.ay.x FT: λx.Ay.(y λx.Ay.x) = λx.Ay.(yT) F²T: Ax.Ay.(y Ax.Ay.(y λx.Ay.x) = x.Ay.(y FT) Fi+¹T: Ax.Ay.(y FT) Now show that: 1.00, 1₁-1 T = Po 2.00, 1-1)FT = ₁ 3.00 1-1 F-1 = -1 Consider the list: (Po, 1₁.n-1) being represented in the λ-calculus as follows: (Po): λx.((x0)4) (Do, 0₁): λx.((x₂)(₁)) (Po, P₁, P₂): λx.((x0) 0₁, 02)) (Po, P1, Pn-1): Ax.((x0) 0₁, 02, 0.1)) is a "dummy" 'end of list' delimiter. It can be any λ-expression. The idea, in Exercise 7.1 on the page before, to let T and F select into a list of length 2 and yield 1st, respectively the second element, is now iterated: T:Ax.ay.x FT: λx.Ay.(y λx.Ay.x) = λx.Ay.(yT) F²T: Ax.Ay.(y Ax.Ay.(y λx.Ay.x) = x.Ay.(y FT) Fi+¹T: Ax.Ay.(y FT) Now show that: 1.00, 1₁-1 T = Po 2.00, 1-1)FT = ₁ 3.00 1-1 F-1 = -1
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