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

  Signature: exp   :: Int -> Int -> Int
             start :: Int

  Rules: exp(x, y) -> x | x > y
         exp(x, y) -> exp(2 * x, y) | x <= y /\ 0 < x /\ x < 100000
         start -> exp(1, y) | 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 UNION R_?, i, c), where:

  P1. (1) exp#(x, y) => exp#(2 * x, y) | x <= y /\ 0 < x /\ x < 100000
      (2) start# => exp#(1, y) | y = y

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 1 

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

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

  P2. (1) exp#(x, y) => exp#(2 * x, y) | x <= y /\ 0 < x /\ x < 100000

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

We use the following integer mapping:

  J(exp#) = 100000 - arg_1 - 1

We thus have:

  (1) x <= y /\ 0 < x /\ x < 100000 |= 100000 - x - 1 > 100000 - 2 * x - 1 (and 100000 - x - 1 >= 0)

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

Processor output: { }.