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

  Signature: cond1 :: Bool -> Int -> Int -> Int
             cond2 :: Bool -> Int -> Int -> Int
             diff  :: Int -> Int -> Int

  Rules: diff(x, y) -> cond1(x = y, x, y) | x = x /\ y = y
         cond1(true, x, y) -> 0 | x = x /\ y = y
         cond1(false, x, y) -> cond2(x > y, x, y) | x = x /\ y = y
         cond2(true, x, y) -> 1 + diff(x, y + 1) | x = x /\ y = y
         cond2(false, x, y) -> 1 + diff(x + 1, y) | 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) diff#(x, y) => cond1#(x = y, x, y) | x = x /\ y = y
      (2) cond1#(false, x, y) => cond2#(x > y, x, y) | x = x /\ y = y
      (3) cond2#(true, x, y) => diff#(x, y + 1) | x = x /\ y = y
      (4) cond2#(false, x, y) => diff#(x + 1, y) | x = x /\ y = y

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

We chain DPs according to the following mapping:


  cond2#(true, x, y) => cond1#(x = y + 1, x, y + 1) | x = x /\ y = y /\ (x = x /\ y + 1 = y + 1)   is obtained by chaining  cond2#(true, x, y) => diff#(x, y + 1) | x = x /\ y = y  and diff#(x', y') => cond1#(x' = y', x', y') | x' = x' /\ y' = y'
  cond2#(false, x, y) => cond1#(x + 1 = y, x + 1, y) | x = x /\ y = y /\ (x + 1 = x + 1 /\ y = y)  is obtained by chaining  cond2#(false, x, y) => diff#(x + 1, y) | x = x /\ y = y and diff#(x', y') => cond1#(x' = y', x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

cond2#(true, x, y) => diff#(x, y + 1) | x = x /\ y = y

cond2#(false, x, y) => diff#(x + 1, y) | x = x /\ y = y

diff#(x, y) => cond1#(x = y, x, y) | x = x /\ y = y


By chaining, we added 2 DPs and removed 3 DPs.

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

  P2. (1) cond1#(false, x, y) => cond2#(x > y, x, y) | x = x /\ y = y
      (2) cond2#(true, x, y) => cond1#(x = y + 1, x, y + 1) | x = x /\ y = y /\ (x = x /\ y + 1 = y + 1)
      (3) cond2#(false, x, y) => cond1#(x + 1 = y, x + 1, y) | x = x /\ y = y /\ (x + 1 = x + 1 /\ y = y)

***** No progress could be made on DP problem D2 = (P2, R, i, c).