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

  Signature: cond  :: Int -> Int -> Int -> Int
             if    :: Bool -> Int -> Int -> Int
             min   :: Int -> Int -> Int
             minus :: Int -> Int -> Int

  Rules: minus(x, x) -> 0 | x = x
         minus(x, y) -> cond(min(x, y), x, y) | x = x /\ y = y
         cond(y, x, y) -> 1 + minus(x, y + 1) | y = y /\ x = x
         min(u, v) -> if(u < v, u, v) | u = u /\ v = v
         if(true, u, v) -> u | u = u /\ v = v
         if(false, u, v) -> v | u = u /\ v = v

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) minus#(x, y) => min#(x, y) | x = x /\ y = y
      (2) minus#(x, y) => cond#(min(x, y), x, y) | x = x /\ y = y
      (3) cond#(y, x, y) => minus#(x, y + 1) | y = y /\ x = x
      (4) min#(u, v) => if#(u < v, u, v) | u = u /\ v = v

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

We compute a graph approximation with the following edges:

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

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) minus#(x, y) => cond#(min(x, y), x, y) | x = x /\ y = y
      (2) cond#(y, x, y) => minus#(x, y + 1) | y = y /\ x = x

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

We chain DPs according to the following mapping:


  cond#(y, x, y) => cond#(min(x, y + 1), x, y + 1) | y = y /\ x = x /\ (x = x /\ y + 1 = y + 1)  is obtained by chaining  cond#(y, x, y) => minus#(x, y + 1) | y = y /\ x = x and minus#(x', y') => cond#(min(x', y'), x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

cond#(y, x, y) => minus#(x, y + 1) | y = y /\ x = x

minus#(x, y) => cond#(min(x, y), 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) cond#(y, x, y) => cond#(min(x, y + 1), x, y + 1) | y = y /\ x = x /\ (x = x /\ y + 1 = y + 1)

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

We obtain 3 usable rules (out of 6 rules in the input problem).

Processor output: { D4 = (P3, R2, i, c) }, where:

  R2. (1) if(true, u, v) -> u | u = u /\ v = v
      (2) if(false, u, v) -> v | u = u /\ v = v
      (3) min(u, v) -> if(u < v, u, v) | u = u /\ v = v

***** No progress could be made on DP problem D4 = (P3, R2, i, c).