solution2012-08-15-v3m
TRANSCRIPT
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8/12/2019 Solution2012-08-15-V3M
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Z,B, . . .
f : Z Z f(x) =
1 x Z x 0
0
g : (Z Z) Z g(h) =
0
h(x)= x Z
t s t s
f S={, 0}
f(S) ={0, 1} f(S) =f(S)
fi(x) = 0 x i
S = {f1, f2, . . .} S = f f(x) = 0 x Z g(S) = = 0 = g(S)
g
(x.42) bot bot bot
42
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> 1
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D f : D D {fi()| i N}
f
D = 2N
D
S D
D
(D, )
f : D D f
f mult nf mult
() n N
n
f
mult
f mult
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S(S (S))S (S Z)
Z
S(S)S Z
S
fi() fi+1()
i N
i= 0 f0() = f1()
i >0
fi1() fi() f
f
f(fi1()) f(fi()) fi() fi+1()
{fi() | i N} D {fi() | i N}
f
f({fi()| i N}) =f({fi()| i N}) (f )
={fi+1
()| i N}=({fi+1()| i N} { })
={fi()| i N}
d
f {fi()| i N} d
fi() d f0() = d
fi() d f f(fi()) f(d) d f fi+1()d
S= {M1, M2, . . .} Mi Mi+1
S=
Mi Mi
Mi
B Mi B e Mi\ B k
e Mk eB Mk B B S
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Mi D
D
Ni := {k N | k i} Ni Ni+1 {N1, N2, . . .}
f(M) =
M
{42}
g
(nf
mult())(x, y) =
0
x= 0 n > 0
x y 0< x < n y=
f mult
g(x, y) =
0
x= 0
x y 0< x y=
h(x, y) =
0
x= 0
x y x= y=
x= (x= 0 y = ) (
)
f f(c1, c2) = f mult(f)(c1, c2)
f(c1, c2) = 0 c1 = 0
c1 = 0 f(c1, c2) =c2+f(c11, c2) c1 =
c1 1 c2 = c1 = 0
c2+f(c1 1, c2) h
c1, c2 Z f(c1, c2) = c2+ f(c1 1, c2)
B H
A (A ) A (E)A (B)
A B
B(E)A H
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Lam
t= fact.(x.(If(LessThanOrEx 1) 1 (Timesx (fact (Minusx 1)))))
= { If True x y.x,
If Falsex y.y,
fix f.f(fixf)}
{ Minus x y z | x, y Z z= x y}
{ Times x y z | x, y Z z= x y}
{ LessThanOrE x y b | x, y Z ((x y b= True) (x > y b= False))}
fixt 1 t
n
f x.fn x
fx.x
fx.fx
Succ n n+ 1 Succ (f x.t) f x.f t t
fn x
Plus
n
m
Succ
n+m m n
(fix(quot x y.If(LessThanx y) 0 (Plus1 (quot (Minusx y) y )))) v w
fixt 1
(f.(f (fixf))) t 1
t (fixt) 1
(x.(If(LessThanOrEx 1) 1 (Timesx (fixt (Minusx 1))))) 1
If(LessThanOrE1 1) 1 (Times1 (fixt (Minus1 1))) ()
If True1 (Times1 (fixt (Minus1 1)))
(x.(y.x)) 1 (Times1 (fixt (Minus1 1)))
(y.1) (Times1 (fixt (Minus1 1)))
1
If if
()
Succ= n f x.f (n f x)
Plus= n m f x.n f (m f x)
Plus= n m.n Succ m
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A0 := {x:: a.a Int} y.y x W
W(A0, y.y x)W(A0+ {y:: b1}, y x)
W(A0+ {y:: b1}, y) = (id,b1)W(A0+ {y:: b1}, x) = (id,b2 Int)
mgu(b1, (b2 Int)b3) = [b1/(b2 Int) b3])= ([b1/(b2 Int) b3], b3)
= ([b1/(b2 Int) b3], ((b2 Int) b3) b3)