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

  Signature: l1    :: Int -> Int -> A        (private)
             l2    :: Int -> Int -> A        (private)
             l3    :: Int -> Int -> Int -> A (private)
             l4    :: Int -> Int -> Int -> A (private)
             l5    :: Int -> Int -> Int -> A (private)
             start :: Int -> A

  Rules: start(x) -> l1(x, 0) | x = x
         l1(x, y) -> l2(x, y) | x > y
         l2(x, y) -> l3(x, y, y) | x = x /\ y = y
         l3(x, y, z) -> l4(x, y, z) | z > 0 /\ x = x /\ y = y
         l4(x, y, z) -> l3(x, y, z - y - 1) | x = x /\ y = y /\ z = z
         l4(x, y, z) -> l5(x, y + 1, z) | x = x /\ y = y /\ z = z
         l5(x, y, z) -> l1(x, y) | x = x /\ y = y /\ z = z

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) start#(x) => l1#(x, 0) | x = x
      (2) l1#(x, y) => l2#(x, y) | x > y                                 (private)
      (3) l2#(x, y) => l3#(x, y, y) | x = x /\ y = y                     (private)
      (4) l3#(x, y, z) => l4#(x, y, z) | z > 0 /\ x = x /\ y = y         (private)
      (5) l4#(x, y, z) => l3#(x, y, z - y - 1) | x = x /\ y = y /\ z = z (private)
      (6) l4#(x, y, z) => l5#(x, y + 1, z) | x = x /\ y = y /\ z = z     (private)
      (7) l5#(x, y, z) => l1#(x, y) | x = x /\ y = y /\ z = z            (private)

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

We compute a graph approximation with the following edges:

  1: 2 
  2: 3 
  3: 4 
  4: 5 6 
  5: 4 
  6: 7 
  7: 2 

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) l1#(x, y) => l2#(x, y) | x > y
      (2) l2#(x, y) => l3#(x, y, y) | x = x /\ y = y
      (3) l3#(x, y, z) => l4#(x, y, z) | z > 0 /\ x = x /\ y = y
      (4) l4#(x, y, z) => l3#(x, y, z - y - 1) | x = x /\ y = y /\ z = z
      (5) l4#(x, y, z) => l5#(x, y + 1, z) | x = x /\ y = y /\ z = z
      (6) l5#(x, y, z) => l1#(x, y) | x = x /\ y = y /\ z = z

***** We apply the Usable Rules Processor on D2 = (P2, R, i, c).

We obtain 0 usable rules (out of 7 rules in the input problem).

Processor output: { D3 = (P2, {}, i, c) }.

***** No progress could be made on DP problem D3 = (P2, {}, i, c).