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

  Signature: comp  :: (Int -> Int) -> (Int -> Int) -> Int -> Int
             fact  :: Int -> (Int -> Int) -> Int
             fact1 :: Int -> Int
             fact2 :: Int -> Int
             id    :: Int -> Int

  Rules: comp(f, g, x) -> f(g(x))
         id(x) -> x
         fact(n, k) -> k(1) | n <= 0
         fact(n, k) -> fact(n - 1, comp(k, [*](n))) | n > 0
         fact1(n) -> fact(n, id)
         fact2(n) -> 1 | n <= 0
         fact2(n) -> n * fact2(n - 1) | n > 0

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_?, f, c), where:

  P1. (1) fact#(n, k) => comp#(k, [*](n), fresh1) | n > 0
      (2) fact#(n, k) => fact#(n - 1, comp(k, [*](n))) | n > 0
      (3) fact1#(n) => id#(fresh1)
      (4) fact1#(n) => fact#(n, id)
      (5) fact2#(n) => fact2#(n - 1) | n > 0

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

We compute a graph approximation with the following edges:

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

There are 2 SCCs.

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

  P2. (1) fact#(n, k) => fact#(n - 1, comp(k, [*](n))) | n > 0

  P3. (1) fact2#(n) => fact2#(n - 1) | n > 0

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

We use the following integer mapping:

  J(fact#) = 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 Integer Function Processor on D3 = (P3, R UNION R_?, f, c).

We use the following integer mapping:

  J(fact2#) = 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: { }.