We consider termination 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, f, 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 Theory Arguments Processor on D1 = (P1, R, f, c).

We use the following theory arguments function:

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

Processor output: { D2 = (P2, R, f, c) ; D3 = (P3, R, f, c) }, where:

  P2. (1) f#(x, y) => if1#(x > y, x, y)
      (2) h#(x, y) => if2#(x > y, x, y) { x, y }
      (3) if1#(true, x, y) => h#(x, y) { x, y }
      (4) if2#(false, x, y) => f#(x, y) { x, y }

  P3. (1) h#(x, y) => if2#(x > y, x, y)
      (2) if1#(true, x, y) => h#(x, y)
      (3) if2#(false, x, y) => f#(x, y)

***** We apply the Theory Arguments Processor on D2 = (P2, R, f, c).

We use the following theory arguments function:

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

Processor output: { D4 = (P4, R, f, c) ; D5 = (P5, R, f, c) }, where:

  P4. (1) f#(x, y) => if1#(x > y, x, y) { x, y }
      (2) h#(x, y) => if2#(x > y, x, y) { x, y }
      (3) if1#(true, x, y) => h#(x, y) { x, y }
      (4) if2#(false, x, y) => f#(x, y) { x, y }

  P5. (1) f#(x, y) => if1#(x > y, x, y)

***** We apply the Graph Processor on D3 = (P3, R, f, c).

We compute a graph approximation with the following edges:

  1: 3 
  2: 1 
  3:

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

Processor output: { }.

***** No progress could be made on DP problem D4 = (P4, R, f, c).