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

We use the following theory arguments function:

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

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

  P2. (1) f#(x) => cond1#(x > 1, x)
      (2) cond1#(true, x) => cond2#(x % 2 = 0, x) { x }
      (3) cond2#(true, x) => f#(x / 2) { x }
      (4) cond2#(false, x) => f#(3 * x + 1) { x }

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

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

We use the following theory arguments function:

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

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

  P4. (1) f#(x) => cond1#(x > 1, x) { x }
      (2) cond1#(true, x) => cond2#(x % 2 = 0, x) { x }
      (3) cond2#(true, x) => f#(x / 2) { x }
      (4) cond2#(false, x) => f#(3 * x + 1) { x }

  P5. (1) f#(x) => cond1#(x > 1, x)

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

We compute a graph approximation with the following edges:

  1: 2 3 
  2:
  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).