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

  Signature: cond1 :: Bool -> Int -> Int
             cond2 :: Bool -> Int -> Int
             f     :: Int -> Int

  Rules: f(x) -> cond1(x > 1, x)
         cond1(true, x) -> cond2(x % 2 = 0, x)
         cond1(false, x) -> x
         cond2(true, x) -> f(x / 2)
         cond2(false, x) -> f(3 * x + 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) f#(x) => cond1#(x > 1, x)
      (2) cond1#(true, x) => cond2#(x % 2 = 0, x)
      (3) cond2#(true, x) => f#(x / 2)
      (4) cond2#(false, x) => f#(3 * x + 1)

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

We chain DPs according to the following mapping:


  cond2#(true, x) => cond1#(x / 2 > 1, x / 2) | true /\ true           is obtained by chaining  cond2#(true, x) => f#(x / 2)      and f#(x') => cond1#(x' > 1, x')
  cond2#(false, x) => cond1#(3 * x + 1 > 1, 3 * x + 1) | true /\ true  is obtained by chaining  cond2#(false, x) => f#(3 * x + 1) and f#(x') => cond1#(x' > 1, x')


The following DPs were deleted:

cond2#(true, x) => f#(x / 2)

cond2#(false, x) => f#(3 * x + 1)

f#(x) => cond1#(x > 1, x)


By chaining, we added 2 DPs and removed 3 DPs.

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

  P2. (1) cond1#(true, x) => cond2#(x % 2 = 0, x)
      (2) cond2#(true, x) => cond1#(x / 2 > 1, x / 2) | true /\ true
      (3) cond2#(false, x) => cond1#(3 * x + 1 > 1, 3 * x + 1) | true /\ true

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

We use the following theory arguments function:

  cond1# : [1, 2]
  cond2# : [1, 2]

Processor output: { D3 = (P3, R UNION R_?, i, c) ; D4 = (P4, R UNION R_?, i, c) }, where:

  P3. (1) cond1#(true, x) => cond2#(x % 2 = 0, x)
      (2) cond2#(true, x) => cond1#(x / 2 > 1, x / 2) | true /\ true /\ x = x
      (3) cond2#(false, x) => cond1#(3 * x + 1 > 1, 3 * x + 1) | true /\ true /\ x = x

  P4. (1) cond2#(true, x) => cond1#(x / 2 > 1, x / 2) | true /\ true
      (2) cond2#(false, x) => cond1#(3 * x + 1 > 1, 3 * x + 1) | true /\ true

***** We apply the Theory Arguments Processor on D3 = (P3, R UNION R_?, i, c).

We use the following theory arguments function:

  cond1# : [1, 2]
  cond2# : [1, 2]

Processor output: { D5 = (P5, R UNION R_?, i, c) ; D6 = (P6, R UNION R_?, i, c) }, where:

  P5. (1) cond1#(true, x) => cond2#(x % 2 = 0, x) | x = x
      (2) cond2#(true, x) => cond1#(x / 2 > 1, x / 2) | true /\ true /\ x = x
      (3) cond2#(false, x) => cond1#(3 * x + 1 > 1, 3 * x + 1) | true /\ true /\ x = x

  P6. (1) cond1#(true, x) => cond2#(x % 2 = 0, x)

***** We apply the Graph Processor on D4 = (P4, R UNION R_?, i, c).

We compute a graph approximation with the following edges:

  1:
  2:

As there are no SCCs, this DP problem is removed.

Processor output: { }.

***** We apply the Usable Rules Processor on D5 = (P5, R UNION R_?, i, c).

We obtain 0 usable rules (out of 5 rules in the input problem).

Processor output: { D7 = (P5, {}, i, c) }.

***** We apply the Graph Processor on D6 = (P6, R UNION R_?, i, c).

We compute a graph approximation with the following edges:

  1:

As there are no SCCs, this DP problem is removed.

Processor output: { }.

***** No progress could be made on DP problem D7 = (P5, {}, i, c).