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

  Signature: if  :: Bool -> Int -> Int -> Int
             pow :: Int -> Int -> Int

  Rules: pow(x, y) -> if(y > 0, x, y) | x = x /\ y = y
         if(true, x, y) -> x * pow(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) pow#(x, y) => if#(y > 0, x, y) | x = x /\ y = y
      (2) if#(true, x, y) => pow#(x, y - 1) | x = x /\ y = y

***** We apply the Chaining Processor Processor on D1 = (P1, 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) => pow#(x, y - 1) | x = x /\ y = y and pow#(x', y') => if#(y' > 0, x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

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

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


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

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

  P2. (1) if#(true, x, y) => if#(y - 1 > 0, x, y - 1) | x = x /\ y = y /\ (x = x /\ y - 1 = y - 1)

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

We obtain 0 usable rules (out of 3 rules in the input problem).

Processor output: { D3 = (P2, {}, i, c) }.

***** No progress could be made on DP problem D3 = (P2, {}, i, c).