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

  Signature: cond  :: Int -> Int -> Int -> Int
             if    :: Bool -> Int -> Int -> Int
             min   :: Int -> Int -> Int
             minus :: Int -> Int -> Int

  Rules: minus(x, x) -> 0
         minus(x, y) -> cond(min(x, y), x, y)
         cond(y, x, y) -> 1 + minus(x, y + 1)
         min(u, v) -> if(u < v, u, v)
         if(true, u, v) -> u
         if(false, 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) minus#(x, y) => min#(x, y)
      (2) minus#(x, y) => cond#(min(x, y), x, y)
      (3) cond#(y, x, y) => minus#(x, y + 1)
      (4) min#(u, v) => if#(u < v, u, v)

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

We compute a graph approximation with the following edges:

  1: 4 
  2: 3 
  3: 1 2 
  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) minus#(x, y) => cond#(min(x, y), x, y)
      (2) cond#(y, x, y) => minus#(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:


  cond#(y, x, y) => cond#(min(x, y + 1), x, y + 1) | true /\ true  is obtained by chaining  cond#(y, x, y) => minus#(x, y + 1) and minus#(x', y') => cond#(min(x', y'), x', y')


The following DPs were deleted:

cond#(y, x, y) => minus#(x, y + 1)

minus#(x, y) => cond#(min(x, y), x, y)


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

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

  P3. (1) cond#(y, x, y) => cond#(min(x, y + 1), x, y + 1) | true /\ true

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

We obtain 3 usable rules (out of 6 rules in the input problem).

Processor output: { D4 = (P3, R2, i, c) }, where:

  R2. (1) if(true, u, v) -> u
      (2) if(false, u, v) -> v
      (3) min(u, v) -> if(u < v, u, v)

***** No progress could be made on DP problem D4 = (P3, R2, i, c).