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

  Signature: eval :: Int -> Int -> o

  Rules: eval(x, y) -> eval(x - 1, z) | x >= 0 /\ z = z
         eval(x, y) -> eval(x, y - 1) | y >= 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) eval#(x, y) => eval#(x - 1, z) | x >= 0 /\ z = z
      (2) eval#(x, y) => eval#(x, y - 1) | y >= 0

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

We use the following theory arguments function:

  eval# : [1, 2]

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

  P2. (1) eval#(x, y) => eval#(x - 1, z) | x >= 0 /\ z = z
      (2) eval#(x, y) => eval#(x, y - 1) | y >= 0 { x, y }

  P3. (1) eval#(x, y) => eval#(x, y - 1) | y >= 0

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

We use the following theory arguments function:

  eval# : [1, 2]

Processor output: { D4 = (P4, R UNION R_?, f, c) ; D5 = (P5, R UNION R_?, f, c) }, where:

  P4. (1) eval#(x, y) => eval#(x - 1, z) | x >= 0 /\ z = z { x, y, z }
      (2) eval#(x, y) => eval#(x, y - 1) | y >= 0 { x, y }

  P5. (1) eval#(x, y) => eval#(x - 1, z) | x >= 0 /\ z = z

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

We use the following integer mapping:

  J(eval#) = arg_2 + 1

We thus have:

  (1) y >= 0 |= y + 1 > y - 1 + 1 (and y + 1 >= 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_?, f, c).

We use the following integer mapping:

  J(eval#) = arg_1

We thus have:

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

We may remove the strictly oriented DPs, which yields:

Processor output: { D6 = (P6, R UNION R_?, f, c) }, where:

  P6. (1) eval#(x, y) => eval#(x, y - 1) | y >= 0 { x, y }

***** We apply the Integer Function Processor on D5 = (P5, R UNION R_?, f, c).

We use the following integer mapping:

  J(eval#) = arg_1 + 1

We thus have:

  (1) x >= 0 /\ z = z |= x + 1 > x - 1 + 1 (and x + 1 >= 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 D6 = (P6, R UNION R_?, f, c).

We use the following integer mapping:

  J(eval#) = arg_2 + 1

We thus have:

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

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

Processor output: { }.