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

  Signature: f :: Int -> A (private)
             g :: Int -> A

  Rules: f(x) -> g(x - 1) | x = x
         g(x) -> f(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 UNION R_?, i, c), where:

  P1. (1) f#(x) => g#(x - 1) | x = x (private)
      (2) g#(x) => f#(x) | x > 0

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

We chain DPs according to the following mapping:


  g#(x) => g#(x - 1) | x > 0 /\ x = x  is obtained by chaining  g#(x) => f#(x) | x > 0 and f#(x') => g#(x' - 1) | x' = x'


The following DPs were deleted:

g#(x) => f#(x) | x > 0

f#(x) => g#(x - 1) | x = x


By chaining, we added 1 DPs and removed 2 DPs.

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

  P2. (1) g#(x) => g#(x - 1) | x > 0 /\ x = x

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

We use the following integer mapping:

  J(g#) = arg_1 - 1

We thus have:

  (1) x > 0 /\ x = x |= x - 1 > x - 1 - 1 (and x - 1 >= 0)

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

Processor output: { }.