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

  Signature: condLoop :: Bool -> Int -> Int -> Int -> Int
             condMod  :: Bool -> Int -> Int -> Int -> Int
             halfExp  :: Int -> Int -> Int -> Int
             pow      :: Int -> Int -> Int
             sqBase   :: Int -> Int -> Int -> Int

  Rules: pow(b, e) -> condLoop(e > 0, b, e, 1) | b = b /\ e = e
         condLoop(false, b, e, r) -> r | b = b /\ e = e /\ r = r
         condLoop(true, b, e, r) -> condMod(e % 2 = 1, b, e, r) | b = b /\ e = e /\ r = r
         condMod(false, b, e, r) -> sqBase(b, e, r) | b = b /\ e = e /\ r = r
         condMod(true, b, e, r) -> sqBase(b, e, r * b) | b = b /\ e = e /\ r = r
         sqBase(b, e, r) -> halfExp(b * b, e, r) | b = b /\ e = e /\ r = r
         halfExp(b, e, r) -> condLoop(e > 0, b, e / 2, r) | b = b /\ e = e /\ r = r

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) pow#(b, e) => condLoop#(e > 0, b, e, 1) | b = b /\ e = e
      (2) condLoop#(true, b, e, r) => condMod#(e % 2 = 1, b, e, r) | b = b /\ e = e /\ r = r
      (3) condMod#(false, b, e, r) => sqBase#(b, e, r) | b = b /\ e = e /\ r = r
      (4) condMod#(true, b, e, r) => sqBase#(b, e, r * b) | b = b /\ e = e /\ r = r
      (5) sqBase#(b, e, r) => halfExp#(b * b, e, r) | b = b /\ e = e /\ r = r
      (6) halfExp#(b, e, r) => condLoop#(e > 0, b, e / 2, r) | b = b /\ e = e /\ r = r

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

We compute a graph approximation with the following edges:

  1: 2 
  2: 3 4 
  3: 5 
  4: 5 
  5: 6 
  6: 2 

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) condLoop#(true, b, e, r) => condMod#(e % 2 = 1, b, e, r) | b = b /\ e = e /\ r = r
      (2) condMod#(false, b, e, r) => sqBase#(b, e, r) | b = b /\ e = e /\ r = r
      (3) condMod#(true, b, e, r) => sqBase#(b, e, r * b) | b = b /\ e = e /\ r = r
      (4) sqBase#(b, e, r) => halfExp#(b * b, e, r) | b = b /\ e = e /\ r = r
      (5) halfExp#(b, e, r) => condLoop#(e > 0, b, e / 2, r) | b = b /\ e = e /\ r = r

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

We obtain 0 usable rules (out of 7 rules in the input problem).

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

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