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

  Signature: condAcc :: Bool -> Int -> Int -> Int
             sqrt    :: Int -> Int
             sqrtAcc :: Int -> Int -> Int

  Rules: sqrt(x) -> sqrtAcc(x, 0)
         sqrtAcc(x, y) -> condAcc(y * y >= x \/ y < 0, x, y)
         condAcc(true, x, y) -> y
         condAcc(false, x, y) -> sqrtAcc(x, y + 1)

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) sqrt#(x) => sqrtAcc#(x, 0)
      (2) sqrtAcc#(x, y) => condAcc#(y * y >= x \/ y < 0, x, y)
      (3) condAcc#(false, x, y) => sqrtAcc#(x, y + 1)

***** 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 
  3: 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) sqrtAcc#(x, y) => condAcc#(y * y >= x \/ y < 0, x, y)
      (2) condAcc#(false, x, y) => sqrtAcc#(x, y + 1)

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

We chain DPs according to the following mapping:


  condAcc#(false, x, y) => condAcc#((y + 1) * (y + 1) >= x \/ y + 1 < 0, x, y + 1) | true /\ true  is obtained by chaining  condAcc#(false, x, y) => sqrtAcc#(x, y + 1) and sqrtAcc#(x', y') => condAcc#(y' * y' >= x' \/ y' < 0, x', y')


The following DPs were deleted:

condAcc#(false, x, y) => sqrtAcc#(x, y + 1)

sqrtAcc#(x, y) => condAcc#(y * y >= x \/ y < 0, x, y)


By chaining, we added 1 DPs and removed 2 DPs.

Processor output: { D3 = (P3, R UNION R_?, i, c) }, where:

  P3. (1) condAcc#(false, x, y) => condAcc#((y + 1) * (y + 1) >= x \/ y + 1 < 0, x, y + 1) | true /\ true

***** We apply the Usable Rules Processor on D3 = (P3, R UNION R_?, i, c).

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

Processor output: { D4 = (P3, {}, i, c) }.

***** No progress could be made on DP problem D4 = (P3, {}, i, c).