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 Usable Rules Processor on D1 = (P1, R, i, c).

We obtain 0 usable rules (out of 5 rules in the input problem).

Processor output: { D2 = (P1, {}, i, c) }.

***** No progress could be made on DP problem D2 = (P1, {}, i, c).