foldLeft in terms of foldRight expansion

gistfile1.scala

def foldRight[A,B](l: List[A], z: B)(f: (A, B) => B): B =
  l match {
    case Nil => z
    case Cons(a, as) => f(a, foldRight(as, z)(f))
  }

def foldLeft[A,B](l: List[A], z: B)(f: (B, A) => B): B =
  foldRight(l, (b:B) => b)((a,g) => b => g(f(b,a)))(z)



foldLeft(List(1,2), 0)(_ + _)

// foldLeft rewritten in terms of foldRight
// A = Int
// B = (Int => Int)
// (_ + _) = f(_,_)
foldRight(List(1,2), (i:Int) => i)((x:Int, g: Int => Int) => (i:Int) => g(f(i,x)))(0)

//---Recurse---

// x = 1
((j:Int) => g(f(j,1)))(0)

// g = foldRight(List(2), (i:Int) => i)((x:Int, g: Int => Int) => (i:Int) => g(f(i,x)))
((j:Int) => foldRight(List(2), (i:Int) => i)((x:Int, g: Int => Int) => (i:Int) => g(f(i,x)))(f(j,1)))(0)

//---Recurse---

// x = 2
((j:Int) => ((k:Int) => g(f(k,2)))(f(j,1)))(0)

//g = foldRight(Nil, (i:Int) => i)((x:Int, g: Int => Int) => (i:Int) => g(f(i,x)))
((j:Int) => ((k:Int) => foldRight(Nil, (i:Int) => i)((x:Int, g: Int => Int) => (i:Int) => g(f(i,x)))(f(k,2)))(f(i,1)))(0)

//---Recurse---

// z = (i:Int) => i
((j:Int) => ((k:Int) => ((i:Int) => i)(f(k,2)))(f(j,1)))(0)

//---Apply and Simplify---

// j = 0
((k:Int) => ((i:Int) => i)(f(k,2)))(f(0,1))

// k = f(0,1)
((i:Int) => i)(f(f(0,1),2))

// i = f(f(0,1),2)
f(f(0,1),2)

// f = (_ + _)
f((0 + 1), 2)

// f = (_ + _)
((0 + 1) + 2)

// looks like foldleft to me!