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

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

  Rules: f(x) -> cond1(x > 1, x) | x = x
         cond1(true, x) -> cond2(x % 2 = 0, x) | x = x
         cond1(false, x) -> x | x = x
         cond2(true, x) -> f(x / 2) | x = x
         cond2(false, x) -> f(3 * x + 1) | x = x

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) f#(x) => cond1#(x > 1, x) | x = x
      (2) cond1#(true, x) => cond2#(x % 2 = 0, x) | x = x
      (3) cond2#(true, x) => f#(x / 2) | x = x
      (4) cond2#(false, x) => f#(3 * x + 1) | x = x

***** 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).