We consider universal computability of the LCSTRS with only rule scheme Calc:

  Signature: cons    :: Int -> intlist -> intlist
             fact    :: Int -> Int
             fact2   :: Int -> Int
             fold    :: (Int -> Int -> Int) -> Int -> intlist -> Int
             genlist :: Int -> intlist
             nil     :: intlist

  Rules: fact(n) -> 1 | n <= 0
         fact(n) -> n * fact2(n - 1) | n > 0
         fact2(n) -> fold([*], 1, genlist(n)) | n = n
         genlist(n) -> nil | n <= 0
         genlist(n) -> cons(n, genlist(n - 1)) | n > 0
         fold(f, y, nil) -> y | y = y
         fold(f, y, cons(x, l)) -> f(x, fold(f, y, l)) | y = y /\ x = x

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R UNION R_?, i, c), where:

  P1. (1) fact#(n) => fact2#(n - 1) | n > 0
      (2) fact2#(n) => genlist#(n) | n = n
      (3) fact2#(n) => fold#([*], 1, genlist(n)) | n = n
      (4) genlist#(n) => genlist#(n - 1) | n > 0
      (5) fold#(f, y, cons(x, l)) => fold#(f, y, l) | y = y /\ x = x

***** We apply the Graph Processor on D1 = (P1, R UNION R_?, i, c).

We compute a graph approximation with the following edges:

  1: 2 3 
  2: 4 
  3: 5 
  4: 4 
  5: 5 

There are 2 SCCs.

Processor output: { D2 = (P2, R UNION R_?, i, c) ; D3 = (P3, R UNION R_?, i, c) }, where:

  P2. (1) genlist#(n) => genlist#(n - 1) | n > 0

  P3. (1) fold#(f, y, cons(x, l)) => fold#(f, y, l) | y = y /\ x = x

***** We apply the Integer Function Processor on D2 = (P2, R UNION R_?, i, c).

We use the following integer mapping:

  J(genlist#) = arg_1

We thus have:

  (1) n > 0 |= n > n - 1 (and n >= 0)

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.

***** We apply the Subterm Criterion Processor on D3 = (P3, R UNION R_?, i, c).

We use the following projection function:

  nu(fold#) = 3

We thus have:

  (1) cons(x, l) |>| l

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.