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

  Signature: eval_1 :: Int -> Int -> Int -> o
             eval_2 :: Int -> Int -> Int -> o

  Rules: eval_1(x, y, z) -> eval_2(x, y, z) | x > z /\ y = y
         eval_2(x, y, z) -> eval_2(x, y - 1, z) | x > z /\ y > z
         eval_2(x, y, z) -> eval_1(x - 1, y, z) | x > z /\ z >= y

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) eval_1#(x, y, z) => eval_2#(x, y, z) | x > z /\ y = y
      (2) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z
      (3) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | x > z /\ z >= y

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

We chain DPs according to the following mapping:


  eval_2#(x, y, z) => eval_2#(x - 1, y, z) | x > z /\ z >= y /\ (x - 1 > z /\ y = y)  is obtained by chaining  eval_2#(x, y, z) => eval_1#(x - 1, y, z) | x > z /\ z >= y and eval_1#(x', y', z') => eval_2#(x', y', z') | x' > z' /\ y' = y'


The following DPs were deleted:

eval_2#(x, y, z) => eval_1#(x - 1, y, z) | x > z /\ z >= y

eval_1#(x, y, z) => eval_2#(x, y, z) | x > z /\ y = y


By chaining, we added 1 DPs and removed 2 DPs.

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

  P2. (1) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z
      (2) eval_2#(x, y, z) => eval_2#(x - 1, y, z) | x > z /\ z >= y /\ (x - 1 > z /\ y = y)

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

We compute a graph approximation with the following edges:

  1: 1 2 
  2: 2 

There are 2 SCCs.

Processor output: { D3 = (P3, R UNION R_?, i, c) ; D4 = (P4, R UNION R_?, i, c) }, where:

  P3. (1) eval_2#(x, y, z) => eval_2#(x - 1, y, z) | x > z /\ z >= y /\ (x - 1 > z /\ y = y)

  P4. (1) eval_2#(x, y, z) => eval_2#(x, y - 1, z) | x > z /\ y > z

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

We use the following integer mapping:

  J(eval_2#) = arg_1 - arg_3

We thus have:

  (1) x > z /\ z >= y /\ (x - 1 > z /\ y = y) |= x - z > x - 1 - z (and x - z >= 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 D4 = (P4, R UNION R_?, i, c).

We use the following integer mapping:

  J(eval_2#) = arg_2 - arg_3

We thus have:

  (1) x > z /\ y > z |= y - z > y - 1 - z (and y - z >= 0)

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

Processor output: { }.