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

  Signature: cu  :: Bool -> Int -> o
             exp :: Int -> Int -> Int
             if  :: Bool -> Int -> Int -> Int

  Rules: cu(true, x) -> cu(x < exp(10, 2), x + 1) | x = x
         exp(x, y) -> if(y > 0, x, y) | x = x /\ y = y
         if(true, x, y) -> x * exp(x, y - 1) | x = x /\ y = y
         if(false, x, y) -> 1 | x = 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) cu#(true, x) => exp#(10, 2) | x = x
      (2) cu#(true, x) => cu#(x < exp(10, 2), x + 1) | x = x
      (3) exp#(x, y) => if#(y > 0, x, y) | x = x /\ y = y
      (4) if#(true, x, y) => exp#(x, y - 1) | x = x /\ y = y

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

We compute a graph approximation with the following edges:

  1: 3 
  2:
  3: 4 
  4: 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) exp#(x, y) => if#(y > 0, x, y) | x = x /\ y = y
      (2) if#(true, x, y) => exp#(x, y - 1) | x = x /\ y = y

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

We chain DPs according to the following mapping:


  if#(true, x, y) => if#(y - 1 > 0, x, y - 1) | x = x /\ y = y /\ (x = x /\ y - 1 = y - 1)  is obtained by chaining  if#(true, x, y) => exp#(x, y - 1) | x = x /\ y = y and exp#(x', y') => if#(y' > 0, x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

if#(true, x, y) => exp#(x, y - 1) | x = x /\ y = y

exp#(x, y) => if#(y > 0, x, y) | x = x /\ y = y


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

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

  P3. (1) if#(true, x, y) => if#(y - 1 > 0, x, y - 1) | x = x /\ y = y /\ (x = x /\ y - 1 = y - 1)

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