We consider termination of the LCTRS with only rule scheme Calc:

  Signature: id_dec :: Int -> Int
             id_inc :: Int -> Int
             rand   :: Int -> Int -> Int
             random :: Int -> Int

  Rules: random(x) -> rand(x, 0) | x = x
         rand(x, y) -> y | x = 0 /\ y = y
         rand(x, y) -> rand(-x + 1, id_inc(y)) | x > 0 /\ y = y
         rand(x, y) -> rand(-x - 1, id_dec(y)) | 0 > x /\ y = y
         id_inc(x) -> x | x = x
         id_inc(x) -> x + 1 | x = x
         id_dec(x) -> x | x = x
         id_dec(x) -> x - 1 | x = x

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, i, c), where:

  P1. (1) random#(x) => rand#(x, 0) | x = x
      (2) rand#(x, y) => id_inc#(y) | x > 0 /\ y = y
      (3) rand#(x, y) => rand#(-x + 1, id_inc(y)) | x > 0 /\ y = y
      (4) rand#(x, y) => id_dec#(y) | 0 > x /\ y = y
      (5) rand#(x, y) => rand#(-x - 1, id_dec(y)) | 0 > x /\ y = y

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

We compute a graph approximation with the following edges:

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

There is only one SCC, so all DPs not inside the SCC can be removed.

Processor output: { D2 = (P2, R, i, c) }, where:

  P2. (1) rand#(x, y) => rand#(-x + 1, id_inc(y)) | x > 0 /\ y = y
      (2) rand#(x, y) => rand#(-x - 1, id_dec(y)) | 0 > x /\ y = y

***** No progress could be made on DP problem D2 = (P2, R, i, c).