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

  Signature: f :: Bool -> Int -> Int -> Int -> o
             g :: Bool -> Int -> Int -> Int -> o

  Rules: f(true, x, y, z) -> g(x > y, x, y, z)
         g(true, x, y, z) -> f(x > z, x, y + 1, z)
         g(true, x, y, z) -> f(x > z, x, y, z + 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_?, f, c), where:

  P1. (1) f#(true, x, y, z) => g#(x > y, x, y, z)
      (2) g#(true, x, y, z) => f#(x > z, x, y + 1, z)
      (3) g#(true, x, y, z) => f#(x > z, x, y, z + 1)

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

We use the following theory arguments function:

  f# : [1, 2, 3, 4]
  g# : [1, 2, 3, 4]

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

  P2. (1) f#(true, x, y, z) => g#(x > y, x, y, z)
      (2) g#(true, x, y, z) => f#(x > z, x, y + 1, z) { x, y, z }
      (3) g#(true, x, y, z) => f#(x > z, x, y, z + 1) { x, y, z }

  P3. (1) g#(true, x, y, z) => f#(x > z, x, y + 1, z)
      (2) g#(true, x, y, z) => f#(x > z, x, y, z + 1)

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

We use the following theory arguments function:

  f# : [1, 2, 3, 4]
  g# : [1, 2, 3, 4]

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

  P4. (1) f#(true, x, y, z) => g#(x > y, x, y, z) { x, y, z }
      (2) g#(true, x, y, z) => f#(x > z, x, y + 1, z) { x, y, z }
      (3) g#(true, x, y, z) => f#(x > z, x, y, z + 1) { x, y, z }

  P5. (1) f#(true, x, y, z) => g#(x > y, x, y, z)

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

We compute a graph approximation with the following edges:

  1:
  2:

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).