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

  Signature: id_dec :: Int -> Int
             id_inc :: Int -> Int
             rand   :: Int -> Int -> Int
             random :: Int -> Int

  Rules: random(x) -> rand(x, 0)
         rand(x, y) -> y | x = 0
         rand(x, y) -> rand(x - 1, id_inc(y)) | x > 0
         rand(x, y) -> rand(x + 1, id_dec(y)) | 0 > x
         id_inc(x) -> x
         id_inc(x) -> x + 1
         id_dec(x) -> x
         id_dec(x) -> x - 1

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) random#(x) => rand#(x, 0)
      (2) rand#(x, y) => id_inc#(y) | x > 0
      (3) rand#(x, y) => rand#(x - 1, id_inc(y)) | x > 0
      (4) rand#(x, y) => id_dec#(y) | 0 > x
      (5) rand#(x, y) => rand#(x + 1, id_dec(y)) | 0 > 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 4 5 
  2:
  3: 2 3 
  4:
  5: 4 5 

There are 2 SCCs.

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

  P2. (1) rand#(x, y) => rand#(x - 1, id_inc(y)) | x > 0

  P3. (1) rand#(x, y) => rand#(x + 1, id_dec(y)) | 0 > x

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

We use the following integer mapping:

  J(rand#) = arg_1

We thus have:

  (1) x > 0 |= x > x - 1 (and x >= 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_?, i, c).

We use the following integer mapping:

  J(rand#) = 0 - arg_1 - 1

We thus have:

  (1) 0 > x |= 0 - x - 1 > 0 - (x + 1) - 1 (and 0 - x - 1 >= 0)

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

Processor output: { }.