We consider universal computability of the LCTRS with only rule scheme Calc:

  Signature: cu  :: Bool -> Int -> o
             exp :: Int -> Int -> Int
             if  :: Bool -> Int -> Int -> Int

  Rules: cu(true, x) -> cu(x < exp(10, 2), x + 1) | x = x
         exp(x, y) -> if(y > 0, x, y) | x = x /\ y = y
         if(true, x, y) -> x * exp(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 UNION R_?, i, c), where:

  P1. (1) cu#(true, x) => exp#(10, 2) | x = x
      (2) cu#(true, x) => cu#(x < exp(10, 2), x + 1) | x = x
      (3) exp#(x, y) => if#(y > 0, x, y) | x = x /\ y = y
      (4) if#(true, x, y) => exp#(x, y - 1) | x = x /\ y = y

***** We apply the Graph Processor on D1 = (P1, R UNION R_?, i, c).

We compute a graph approximation with the following edges:

  1: 3 
  2:
  3: 4 
  4: 3 

There is only one SCC, so all DPs not inside the SCC can be removed.

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

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

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

We obtain 0 usable rules (out of 4 rules in the input problem).

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

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