[Functional Programming] Add, Mult, Pow, isZero

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const log = console.log;   // zero :: &fa.a const zero = f => x => x; // zero is F // once :: &fa.fa const once = f => x => f(x); // once it I // twice :: &fa.f(fa) const twice = f => x => f(f(x)); // thrice :: &fa.f(f(fa)) const thrice = f => x => f(f(f(x)));   const T = true; const F = false; const I = x => x; const not = x => !x; const K = x => y => x   log(zero(not)(T)) // true, because only return second arguement log(once(not)(T)) // false log(twice(not)(F)) // false log(thrice(not)(T)) // false   log('****')   /** SUCCSOR SUCC N1 = N2 SUCC N2 = N3 SUCC(SUCC N1) = N3   SUCC &fa.fa = &fa.f(fa) SUCC N2, then n is 2, do f n times, then add one f more */ const _succ = n => f => x => f(n(f)(x)); // conver chunch number to JS number. // jsnum :: take a chunch number, call (x => x + 1) n times, and start from 0. const jsnum = n => n(x => x + 1)(0); log(_succ(zero)(not)(T)) // false log(jsnum(_succ(zero))) // 1 log(jsnum(_succ(_succ(zero)))) // 2   const n0 = zero; const n1 = once; const n2 = twice; const n3 = thrice; const n4 = _succ(thrice);   log(jsnum(_succ(n2))) // 3   const B = f => g => a => f(g(a));   const succ = n => f => B(f)(n(f)); // Add N1 N4 = succ(N4) // Add N2 N4 = succ(succ(N4)) // Add N3 N4 = succ(succ(succ(N4))) // Add N3 N4 = (succ.succ.succ) N4 === N3 succ N4 const add = n => k => n(succ)(k); console.log(jsnum(add(n3)(n4))); // 7   const mult = B; // mult = B console.log(jsnum(mult(n2)(n3)))   // Thrush $af.fa = CI (Cardinal Idiot, flip the arguements) const pow = n => k => k(n); console.log(jsnum(pow(n2)(n3))); // 8   // isZero :: $n.n(f)(args) // is n = 0, f won't run, just return args // Then args should be T // $n.n(f)(T), now if n > 0, f will be run, // we want it always return F // K(F), constant(F) // $n.n(K(F))(T) const isZero = n => n(K(F))(T) console.log(isZero(n0)) // true console.log(isZero(n1)) // false

　succ :: Doing N + 1 times fn.

   add :: Doing N times succ, based on K

   mult :: is B

   pow :: or Thrush, is flip

   isZero :: return just T otherwise K(F) , K is constant