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(i, j) -> eval_2(i, 0) | i >= 0
         eval_2(i, j) -> eval_2(i, j + 1) | j <= i - 1
         eval_2(i, j) -> eval_1(i - 1, j) | j > i - 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) eval_1#(i, j) => eval_2#(i, 0) | i >= 0
      (2) eval_2#(i, j) => eval_2#(i, j + 1) | j <= i - 1
      (3) eval_2#(i, j) => eval_1#(i - 1, j) | j > i - 1

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

We chain DPs according to the following mapping:


  eval_2#(i, j) => eval_2#(i - 1, 0) | j > i - 1 /\ i - 1 >= 0  is obtained by chaining  eval_2#(i, j) => eval_1#(i - 1, j) | j > i - 1 and eval_1#(i', j') => eval_2#(i', 0) | i' >= 0


The following DPs were deleted:

eval_2#(i, j) => eval_1#(i - 1, j) | j > i - 1

eval_1#(i, j) => eval_2#(i, 0) | i >= 0


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

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

  P2. (1) eval_2#(i, j) => eval_2#(i, j + 1) | j <= i - 1
      (2) eval_2#(i, j) => eval_2#(i - 1, 0) | j > i - 1 /\ i - 1 >= 0

***** 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) j <= i - 1              |= i >= i
  (2) j > i - 1 /\ i - 1 >= 0 |= i >  i - 1 (and i >= 0)

We may remove the strictly oriented DPs, which yields:

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

  P3. (1) eval_2#(i, j) => eval_2#(i, j + 1) | j <= i - 1

***** 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 - 1 - arg_2

We thus have:

  (1) j <= i - 1 |= i - 1 - j > i - 1 - (j + 1) (and i - 1 - j >= 0)

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

Processor output: { }.