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

  Signature: cond  :: Bool -> Int -> Int -> Int
             minus :: Int -> Int -> Int

  Rules: minus(x, y) -> cond(x >= y + 1, x, y) | x = x /\ y = y
         cond(false, x, y) -> 0 | x = x /\ y = y
         cond(true, x, y) -> 1 + minus(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 UNION R_?, i, c), where:

  P1. (1) minus#(x, y) => cond#(x >= y + 1, x, y) | x = x /\ y = y
      (2) cond#(true, x, y) => minus#(x, y + 1) | x = x /\ y = y

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

We chain DPs according to the following mapping:


  cond#(true, x, y) => cond#(x >= y + 1 + 1, x, y + 1) | x = x /\ y = y /\ (x = x /\ y + 1 = y + 1)  is obtained by chaining  cond#(true, x, y) => minus#(x, y + 1) | x = x /\ y = y and minus#(x', y') => cond#(x' >= y' + 1, x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

cond#(true, x, y) => minus#(x, y + 1) | x = x /\ y = y

minus#(x, y) => cond#(x >= y + 1, x, y) | x = x /\ y = y


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

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

  P2. (1) cond#(true, x, y) => cond#(x >= y + 1 + 1, x, y + 1) | x = x /\ y = y /\ (x = x /\ y + 1 = y + 1)

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

We obtain 0 usable rules (out of 3 rules in the input problem).

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

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