We consider termination of the LCSTRS with only rule scheme Calc:

  Signature: myfun   :: Int -> Int -> Int
             readint :: Int
             rec     :: (Int -> Int -> Int) -> Int -> Int -> Int

  Rules: rec(F, x, y) -> y | x <= 0
         rec(F, x, y) -> F(x, rec(F, x - 1, y)) | x > 0
         readint -> y | y = y
         myfun(x, z) -> x + z

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) rec#(F, x, y) => rec#(F, x - 1, y) | x > 0

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

We use the following integer mapping:

  J(rec#) = arg_2

We thus have:

  (1) x > 0 |= x > x - 1 (and x >= 0)

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.