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

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

  Rules: eval_1(x, y) -> eval_2(x + 1, 1) | x >= 0 /\ y = y
         eval_2(x, y) -> eval_2(x, y + 1) | x >= 0 /\ y > 0 /\ x >= y
         eval_2(x, y) -> eval_1(x - 2, y) | x >= 0 /\ y > 0 /\ y > 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) eval_1#(x, y) => eval_2#(x + 1, 1) | x >= 0 /\ y = y
      (2) eval_2#(x, y) => eval_2#(x, y + 1) | x >= 0 /\ y > 0 /\ x >= y
      (3) eval_2#(x, y) => eval_1#(x - 2, y) | x >= 0 /\ y > 0 /\ y > x

***** 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) => eval_2#(x - 2 + 1, 1) | x >= 0 /\ y > 0 /\ y > x /\ (x - 2 >= 0 /\ y = y)  is obtained by chaining  eval_2#(x, y) => eval_1#(x - 2, y) | x >= 0 /\ y > 0 /\ y > x and eval_1#(x', y') => eval_2#(x' + 1, 1) | x' >= 0 /\ y' = y'


The following DPs were deleted:

eval_2#(x, y) => eval_1#(x - 2, y) | x >= 0 /\ y > 0 /\ y > x

eval_1#(x, y) => eval_2#(x + 1, 1) | x >= 0 /\ 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) => eval_2#(x, y + 1) | x >= 0 /\ y > 0 /\ x >= y
      (2) eval_2#(x, y) => eval_2#(x - 2 + 1, 1) | x >= 0 /\ y > 0 /\ y > x /\ (x - 2 >= 0 /\ y = y)

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

We use the following integer mapping:

  J(eval_2#) = arg_1

We thus have:

  (1) x >= 0 /\ y > 0 /\ x >= y                         |= x >= x
  (2) x >= 0 /\ y > 0 /\ y > x /\ (x - 2 >= 0 /\ y = y) |= x >  x - 2 + 1 (and x >= 0)

We may remove the strictly oriented DPs, which yields:

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

  P3. (1) eval_2#(x, y) => eval_2#(x, y + 1) | x >= 0 /\ y > 0 /\ x >= y

***** 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_2

We thus have:

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

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

Processor output: { }.