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

  Signature: f :: Bool -> Int -> Int -> Int -> o
             g :: Bool -> Int -> Int -> Int -> o

  Rules: f(true, x, y, z) -> g(x > y, x, y, z) | x = x /\ y = y /\ z = z
         g(true, x, y, z) -> f(x > z, x, y + 1, z) | x = x /\ y = y /\ z = z
         g(true, x, y, z) -> f(x > z, x, y, z + 1) | x = x /\ y = y /\ z = z

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#(true, x, y, z) => g#(x > y, x, y, z) | x = x /\ y = y /\ z = z
      (2) g#(true, x, y, z) => f#(x > z, x, y + 1, z) | x = x /\ y = y /\ z = z
      (3) g#(true, x, y, z) => f#(x > z, x, y, z + 1) | x = x /\ y = y /\ z = z

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

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

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

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