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

  Signature: ack    :: Int -> Int -> Int
             ackNat :: Bool -> Int -> Int -> Int
             cond1  :: Bool -> Int -> Int -> Int
             cond2  :: Bool -> Int -> Int -> Int
             f      :: Bool -> Int -> o

  Rules: f(true, x) -> f(ack(10, 10) >= x, x + 1) | x = x
         ack(x, y) -> ackNat(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
         ackNat(true, x, y) -> cond1(x = 0, x, y) | x = x /\ y = y
         ackNat(false, x, y) -> 0 | x = x /\ y = y
         cond1(true, x, y) -> y + 1 | x = x /\ y = y
         cond1(false, x, y) -> cond2(y = 0, x, y) | x = x /\ y = y
         cond2(true, x, y) -> ack(x - 1, y + 1) | x = x /\ y = y
         cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) | x = x /\ y = 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, i, c), where:

  P1. (1) f#(true, x) => ack#(10, 10) | x = x
      (2) f#(true, x) => f#(ack(10, 10) >= x, x + 1) | x = x
      (3) ack#(x, y) => ackNat#(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
      (4) ackNat#(true, x, y) => cond1#(x = 0, x, y) | x = x /\ y = y
      (5) cond1#(false, x, y) => cond2#(y = 0, x, y) | x = x /\ y = y
      (6) cond2#(true, x, y) => ack#(x - 1, y + 1) | x = x /\ y = y
      (7) cond2#(false, x, y) => ack#(x, y - 1) | x = x /\ y = y
      (8) cond2#(false, x, y) => ack#(x - 1, ack(x, y - 1)) | x = x /\ y = y

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

We compute a graph approximation with the following edges:

  1: 3 
  2:
  3: 4 
  4: 5 
  5: 6 7 8 
  6: 3 
  7: 3 
  8: 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) ack#(x, y) => ackNat#(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
      (2) ackNat#(true, x, y) => cond1#(x = 0, x, y) | x = x /\ y = y
      (3) cond1#(false, x, y) => cond2#(y = 0, x, y) | x = x /\ y = y
      (4) cond2#(true, x, y) => ack#(x - 1, y + 1) | x = x /\ y = y
      (5) cond2#(false, x, y) => ack#(x, y - 1) | x = x /\ y = y
      (6) cond2#(false, x, y) => ack#(x - 1, ack(x, y - 1)) | x = x /\ y = y

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

We obtain 7 usable rules (out of 8 rules in the input problem).

Processor output: { D3 = (P2, R2, i, c) }, where:

  R2. (1) ackNat(true, x, y) -> cond1(x = 0, x, y) | x = x /\ y = y
      (2) ackNat(false, x, y) -> 0 | x = x /\ y = y
      (3) ack(x, y) -> ackNat(x >= 0 /\ y >= 0, x, y) | x = x /\ y = y
      (4) cond1(true, x, y) -> y + 1 | x = x /\ y = y
      (5) cond1(false, x, y) -> cond2(y = 0, x, y) | x = x /\ y = y
      (6) cond2(true, x, y) -> ack(x - 1, y + 1) | x = x /\ y = y
      (7) cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) | x = x /\ y = y

***** No progress could be made on DP problem D3 = (P2, R2, i, c).