lambda-calculus-with-js/idea1.txt

Using JS' anonymous functions to perform lambda calculus

2λa 0 b 0
3(a 0 b) 0
1a 1b 1c 1d

lambda = sign | func | call
sign = (1x)+ 0
func = 2 lambda
call = 3 lambda lambda

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(λa.a b)
3	2	100	100	110
130	120	1110100	1110100	1111100
2 130	2 120	2 a	2 a	2 b
c f 1 1 b

c	λz.λw.(P (z w)) = λz.λw.(z w)
f	λx.(P λa.x) = λx.(λz.λw.(z w) λa.x) = λx.λw.(λa.x w)
a	(P a) = (λx.λw.(λa.x w) a) = λw.(λa.a w)
a	(P a) = (λw.(λa.a w) a) = λa.a
b	(P b) = (λa.a b)

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λa.λb.(a b)
f f c 0 1

f a	λx.(P λa.x) = λx.(λn.n λa.x) = λx.λa.x
f	λy.(P λi.y) = λx.λa.x
f b	λy.(P λb.y) = λy.(λx.λa.x λb.y) = λy.λa.λb.y
c	λz.λw.(P (z w)) = λz.λw.(λy.λa.λb.y (z w)) = λz.λw.λa.λb.(z w)
a	(P a) = (λz.λw.λa.λb.(z w) a) = λw.λa.λb.(a w)
b	(P b) = (λw.λa.λb.(a w) b) = λa.λb.(a b)

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true = \x.\y.x
false = \x.\y.y
and = \p.\q.p q p
or = \p.\q.p p q
not = \p.p false true
Yc = λp.(λf.λa.((p (f f)) a))(λf.λa.((p (f f)) a))
0 = \f.\x.x
1 = \f.\x.(f x)
2 = \f.\x.(f (f x))
3 = \f.\x.(f (f (f x)))
\a.\b.b


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L0 = \T.T \F.F 0
L1 = \T.T \F.(F a) 1
L2 = \T.T \F.((F a) b) 2
L3 = \T.T \F.(((F a) b) c) 3
\F.(((F a) b) c)


push_head = \L.\x.\T.T (L former (F x)) (succ (L latter))
push_tail = \L.\x.\F.((L F) x)
delete_head = \L.\F.(L \x.F)
#reverse = \L.\F.(L )
n+1_params_fn = \n.(n (λa.λb.a))
many_params_fn = ((Y (\s.\r.\a.(s (succ r)))) 0)

\a.(a \b.(b a))
f c 0 f c 1 0

f	\A.			\a.(A [a])
c	\B.\C.		\a.((B [a]) (C [a]))	c = \P.\X.\Y.(P \L.((X L) (Y L)))
0	\C.			\a.(a (C [a]))			0 = \P.(P \L.getL L 0)
f	\D.			\a.(a \b.(D [a,b]))		f = \P.\D.(P \L.\x.(D (push L x)))
c	\E.\F.		\a.(a \b.((E [a,b]) (F [a,b])))
1	\F.			\a.(a \b.(b (F [a,b])))
0				\a.(a \b.(b a))

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(\a.(\b.b a) \a.a)
c f c f 1 0 f 0

c	\A.\B.		((A []) (B []))
f	\C.\B.		(\a.(C [a]) (B []))
c	\D.\E.\B.	(\a.((D [a]) (E [a])) (B []))
f	\F.\E.\B.	(\a.(\b.(F [a,b]) (E [a])) (B []))
1	\E.\B.		(\a.(\b.b (E [a])) (B []))
0	\B.			(\a.(\b.b a) (B []))
f	\G.			(\a.(\b.b a) (\a.(G [a])))
0				(\a.(\b.b a) (\a.a))

f = λP.λx.(P λa.x)
c = λP.λz.λw.(P (z w))
* = λP.(P *)

a = λs.λr.λn.(s (n r))



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l = λf.(((f a) b) c)
l[0] = (l λa.λb.λc.a) = (λf.(((f a) b) c) λa.λb.λc.a)


2 = \f.\x.(f (f x))
plus = \a.\b.\F.\X.((a F) ((b F) X))
succ = \a.\F.\X.((a F) (F X))

\g.\h.(h (g F))
2 = \h.(h \F.(h X))