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

  Signature: div :: Int -> Int -> Int
             if  :: Bool -> Int -> Int -> Int

  Rules: div(x, y) -> if(x >= y /\ y > 0, x, y)
         if(true, x, y) -> div(x - y, y) + 1

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) div#(x, y) => if#(x >= y /\ y > 0, x, y)
      (2) if#(true, x, y) => div#(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:


  if#(true, x, y) => if#(x - y >= y /\ y > 0, x - y, y) | true /\ true  is obtained by chaining  if#(true, x, y) => div#(x - y, y) and div#(x', y') => if#(x' >= y' /\ y' > 0, x', y')


The following DPs were deleted:

if#(true, x, y) => div#(x - y, y)

div#(x, y) => if#(x >= y /\ y > 0, x, y)


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

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

  P2. (1) if#(true, x, y) => if#(x - y >= y /\ y > 0, x - y, y) | true /\ true

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

We obtain 0 usable rules (out of 2 rules in the input problem).

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

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