We consider termination 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 > y /\ z = z
         eval_2(x, y, z) -> eval_1(x, y + 1, z) | x > z /\ y = y
         eval_2(x, y, z) -> eval_1(x, y, z + 1) | x > z /\ y = y
         eval_2(x, y, z) -> eval_1(x - 1, y, z) | z >= x /\ y = 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, i, c), where:

  P1. (1) eval_1#(x, y, z) => eval_2#(x, y, z) | x > y /\ z = z
      (2) eval_2#(x, y, z) => eval_1#(x, y + 1, z) | x > z /\ y = y
      (3) eval_2#(x, y, z) => eval_1#(x, y, z + 1) | x > z /\ y = y
      (4) eval_2#(x, y, z) => eval_1#(x - 1, y, z) | z >= x /\ y = y

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

We chain DPs according to the following mapping:


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


The following DPs were deleted:

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

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

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

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


By chaining, we added 3 DPs and removed 4 DPs.

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

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

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

We compute a graph approximation with the following edges:

  1: 1 2 
  2: 1 2 3 
  3: 3 

There are 2 SCCs.

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

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

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

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

We use the following integer mapping:

  J(eval_2#) = arg_1 - 1 - arg_2

We thus have:

  (1) z >= x /\ y = y /\ (x - 1 > y /\ z = z) |= x - 1 - y > x - 1 - 1 - y (and x - 1 - y >= 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, i, c).

We use the following integer mapping:

  J(eval_2#) = arg_1 - (arg_2 + 1)

We thus have:

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

We may remove the strictly oriented DPs, which yields:

Processor output: { D5 = (P5, R, i, c) }, where:

  P5. (1) eval_2#(x, y, z) => eval_2#(x, y, z + 1) | x > z /\ y = y /\ (x > y /\ z + 1 = z + 1)

***** We apply the Integer Function Processor on D5 = (P5, R, i, c).

We use the following integer mapping:

  J(eval_2#) = arg_1 - arg_3 - 1

We thus have:

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

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

Processor output: { }.