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

  Signature: f   :: Int -> Int -> Int
             h   :: Int -> Int -> Int
             if1 :: Bool -> Int -> Int -> Int
             if2 :: Bool -> Int -> Int -> Int

  Rules: f(x, y) -> if1(x > y, x, y)
         h(x, y) -> if2(x > y, x, y)
         if1(true, x, y) -> h(x, y)
         if1(false, x, y) -> 0
         if2(true, x, y) -> 0
         if2(false, x, y) -> f(x, 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) f#(x, y) => if1#(x > y, x, y)
      (2) h#(x, y) => if2#(x > y, x, y)
      (3) if1#(true, x, y) => h#(x, y)
      (4) if2#(false, x, y) => f#(x, y)

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

We chain DPs according to the following mapping:


  if2#(false, x, y) => if1#(x > y, x, y) | true /\ true  is obtained by chaining  if2#(false, x, y) => f#(x, y) and f#(x', y') => if1#(x' > y', x', y')


The following DPs were deleted:

if2#(false, x, y) => f#(x, y)

f#(x, y) => if1#(x > y, x, y)


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

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

  P2. (1) h#(x, y) => if2#(x > y, x, y)
      (2) if1#(true, x, y) => h#(x, y)
      (3) if2#(false, x, y) => if1#(x > y, x, y) | true /\ true

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

We chain DPs according to the following mapping:


  if1#(true, x, y) => if2#(x > y, x, y) | true /\ true  is obtained by chaining  if1#(true, x, y) => h#(x, y) and h#(x', y') => if2#(x' > y', x', y')


The following DPs were deleted:

if1#(true, x, y) => h#(x, y)

h#(x, y) => if2#(x > y, x, y)


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

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

  P3. (1) if2#(false, x, y) => if1#(x > y, x, y) | true /\ true
      (2) if1#(true, x, y) => if2#(x > y, x, y) | true /\ true

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

We use the following theory arguments function:

  if1# : [1, 2, 3]
  if2# : [1, 2, 3]

Processor output: { D4 = (P4, R UNION R_?, i, c) ; D5 = (P5, R UNION R_?, i, c) }, where:

  P4. (1) if2#(false, x, y) => if1#(x > y, x, y) | true /\ true
      (2) if1#(true, x, y) => if2#(x > y, x, y) | true /\ true /\ x = x /\ y = y

  P5. (1) if1#(true, x, y) => if2#(x > y, x, y) | true /\ true

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

We use the following theory arguments function:

  if1# : [1, 2, 3]
  if2# : [1, 2, 3]

Processor output: { D6 = (P6, R UNION R_?, i, c) ; D7 = (P7, R UNION R_?, i, c) }, where:

  P6. (1) if2#(false, x, y) => if1#(x > y, x, y) | true /\ true /\ x = x /\ y = y
      (2) if1#(true, x, y) => if2#(x > y, x, y) | true /\ true /\ x = x /\ y = y

  P7. (1) if2#(false, x, y) => if1#(x > y, x, y) | true /\ true

***** We apply the Graph Processor on D5 = (P5, 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: { }.

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

We obtain 0 usable rules (out of 6 rules in the input problem).

Processor output: { D8 = (P6, {}, i, c) }.

***** We apply the Graph Processor on D7 = (P7, 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 D8 = (P6, {}, i, c).