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

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, i, c), where:

  P1. (1) f#(x) => cond1#(x > 1, x) | x = x
      (2) cond1#(true, x) => cond2#(x % 2 = 0, x) | x = x
      (3) cond2#(true, x) => f#(x / 2) | x = x
      (4) cond2#(false, x) => f#(3 * x + 1) | x = x

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

We chain DPs according to the following mapping:


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


The following DPs were deleted:

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

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

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


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

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

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

***** No progress could be made on DP problem D2 = (P2, R, i, c).