We consider termination of the LCTRS with only rule scheme Calc:

  Signature: cond   :: Bool -> Int -> Int -> Int -> Int
             d      :: Int -> Int -> Int -> Int
             dNat   :: Bool -> Int -> Int -> Int -> Int
             div    :: Int -> Int -> Int
             divNat :: Bool -> Int -> Int -> Int

  Rules: div(x, y) -> divNat(x >= 0 /\ y >= 1, x, y) | x = x /\ y = y
         divNat(true, x, y) -> d(x, y, 0) | x = x /\ y = y
         d(x, y, z) -> dNat(x >= 0 /\ y >= 1 /\ z >= 0, x, y, z) | x = x /\ y = y /\ z = z
         dNat(true, x, y, z) -> cond(x >= z, x, y - 1, z) | x = x /\ y = y /\ z = z
         cond(true, x, y, z) -> 1 + d(x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z
         cond(false, x, y, z) -> 0 | x = x /\ y = y /\ z = z

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) div#(x, y) => divNat#(x >= 0 /\ y >= 1, x, y) | x = x /\ y = y
      (2) divNat#(true, x, y) => d#(x, y, 0) | x = x /\ y = y
      (3) d#(x, y, z) => dNat#(x >= 0 /\ y >= 1 /\ z >= 0, x, y, z) | x = x /\ y = y /\ z = z
      (4) dNat#(true, x, y, z) => cond#(x >= z, x, y - 1, z) | x = x /\ y = y /\ z = z
      (5) cond#(true, x, y, z) => d#(x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z

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

We compute a graph approximation with the following edges:

  1: 2 
  2: 3 
  3: 4 
  4: 5 
  5: 3 

There is only one SCC, so all DPs not inside the SCC can be removed.

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

  P2. (1) d#(x, y, z) => dNat#(x >= 0 /\ y >= 1 /\ z >= 0, x, y, z) | x = x /\ y = y /\ z = z
      (2) dNat#(true, x, y, z) => cond#(x >= z, x, y - 1, z) | x = x /\ y = y /\ z = z
      (3) cond#(true, x, y, z) => d#(x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z

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

We chain DPs according to the following mapping:


  cond#(true, x, y, z) => dNat#(x >= 0 /\ y + 1 >= 1 /\ y + 1 + z >= 0, x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z /\ (x = x /\ y + 1 = y + 1 /\ y + 1 + z = y + 1 + z)  is obtained by chaining  cond#(true, x, y, z) => d#(x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z and d#(x', y', z') => dNat#(x' >= 0 /\ y' >= 1 /\ z' >= 0, x', y', z') | x' = x' /\ y' = y' /\ z' = z'


The following DPs were deleted:

cond#(true, x, y, z) => d#(x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z

d#(x, y, z) => dNat#(x >= 0 /\ y >= 1 /\ z >= 0, x, y, z) | x = x /\ y = y /\ z = z


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

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

  P3. (1) dNat#(true, x, y, z) => cond#(x >= z, x, y - 1, z) | x = x /\ y = y /\ z = z
      (2) cond#(true, x, y, z) => dNat#(x >= 0 /\ y + 1 >= 1 /\ y + 1 + z >= 0, x, y + 1, y + 1 + z) | x = x /\ y = y /\ z = z /\ (x = x /\ y + 1 = y + 1 /\ y + 1 + z = y + 1 + z)

***** No progress could be made on DP problem D3 = (P3, R, i, c).