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

  Signature: cons    :: o -> Int -> o
             conv    :: Int -> o
             lastbit :: Int -> Int
             nil     :: o

  Rules: lastbit(0) -> 0
         lastbit(1) -> 1
         lastbit(x) -> lastbit(x - 2) | x > 1
         conv(0) -> cons(nil, 0)
         conv(x) -> cons(conv(x / 2), lastbit(x)) | x > 0

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) lastbit#(x) => lastbit#(x - 2) | x > 1
      (2) conv#(x) => conv#(x / 2) | x > 0
      (3) conv#(x) => lastbit#(x) | x > 0

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 2 3 
  3: 1 

There are 2 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) }, where:

  P2. (1) lastbit#(x) => lastbit#(x - 2) | x > 1

  P3. (1) conv#(x) => conv#(x / 2) | x > 0

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

We use the following integer mapping:

  J(lastbit#) = arg_1

We thus have:

  (1) x > 1 |= x > x - 2 (and x >= 0)

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.

***** We apply the Integer Function Processor on D3 = (P3, R, i, c).

We use the following integer mapping:

  J(conv#) = arg_1

We thus have:

  (1) x > 0 |= x > x / 2 (and x >= 0)

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.