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

  Signature: f     :: Bool -> Int -> Int -> o
             fNat  :: Bool -> Int -> Int -> o
             if    :: Bool -> Int -> Int -> Int
             round :: Int -> Int

  Rules: f(true, x, y) -> fNat(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
         fNat(true, x, y) -> f(x > y, x, round(y + 1)) | x = x /\ y = y
         round(x) -> if(x % 2 = 0, x, x + 1) | x = x
         if(true, u, v) -> u | u = u /\ v = v
         if(false, u, v) -> v | u = u /\ v = v

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) f#(true, x, y) => fNat#(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
      (2) fNat#(true, x, y) => round#(y + 1) | x = x /\ y = y
      (3) fNat#(true, x, y) => f#(x > y, x, round(y + 1)) | x = x /\ y = y
      (4) round#(x) => if#(x % 2 = 0, x, x + 1) | x = x

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

We compute a graph approximation with the following edges:

  1: 2 3 
  2: 4 
  3: 1 
  4:

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) f#(true, x, y) => fNat#(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
      (2) fNat#(true, x, y) => f#(x > y, x, round(y + 1)) | x = x /\ y = y

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

We obtain 3 usable rules (out of 5 rules in the input problem).

Processor output: { D3 = (P2, R2, i, c) }, where:

  R2. (1) if(true, u, v) -> u | u = u /\ v = v
      (2) if(false, u, v) -> v | u = u /\ v = v
      (3) round(x) -> if(x % 2 = 0, x, x + 1) | x = x

***** No progress could be made on DP problem D3 = (P2, R2, i, c).