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

  Signature: cond1 :: Bool -> Int -> Int -> Int
             cond2 :: Bool -> Int -> Int -> Int
             diff  :: Int -> Int -> Int

  Rules: diff(x, y) -> cond1(x = y, x, y)
         cond1(true, x, y) -> 0
         cond1(false, x, y) -> cond2(x > y, x, y)
         cond2(true, x, y) -> 1 + diff(x, y + 1)
         cond2(false, x, y) -> 1 + diff(x + 1, 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_?, f, c), where:

  P1. (1) diff#(x, y) => cond1#(x = y, x, y)
      (2) cond1#(false, x, y) => cond2#(x > y, x, y)
      (3) cond2#(true, x, y) => diff#(x, y + 1)
      (4) cond2#(false, x, y) => diff#(x + 1, y)

***** We apply the Theory Arguments Processor on D1 = (P1, R UNION R_?, f, c).

We use the following theory arguments function:

  cond1# : [1, 2, 3]
  cond2# : [1, 2, 3]
  diff#  : [1, 2]

Processor output: { D2 = (P2, R UNION R_?, f, c) ; D3 = (P3, R UNION R_?, f, c) }, where:

  P2. (1) diff#(x, y) => cond1#(x = y, x, y)
      (2) cond1#(false, x, y) => cond2#(x > y, x, y) { x, y }
      (3) cond2#(true, x, y) => diff#(x, y + 1) { x, y }
      (4) cond2#(false, x, y) => diff#(x + 1, y) { x, y }

  P3. (1) cond1#(false, x, y) => cond2#(x > y, x, y)
      (2) cond2#(true, x, y) => diff#(x, y + 1)
      (3) cond2#(false, x, y) => diff#(x + 1, y)

***** We apply the Theory Arguments Processor on D2 = (P2, R UNION R_?, f, c).

We use the following theory arguments function:

  cond1# : [1, 2, 3]
  cond2# : [1, 2, 3]
  diff#  : [1, 2]

Processor output: { D4 = (P4, R UNION R_?, f, c) ; D5 = (P5, R UNION R_?, f, c) }, where:

  P4. (1) diff#(x, y) => cond1#(x = y, x, y) { x, y }
      (2) cond1#(false, x, y) => cond2#(x > y, x, y) { x, y }
      (3) cond2#(true, x, y) => diff#(x, y + 1) { x, y }
      (4) cond2#(false, x, y) => diff#(x + 1, y) { x, y }

  P5. (1) diff#(x, y) => cond1#(x = y, x, y)

***** We apply the Graph Processor on D3 = (P3, R UNION R_?, f, c).

We compute a graph approximation with the following edges:

  1: 2 3 
  2:
  3:

As there are no SCCs, this DP problem is removed.

Processor output: { }.

***** No progress could be made on DP problem D4 = (P4, R UNION R_?, f, c).