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

  Signature: cond   :: Bool -> Int -> Int -> Int
             log    :: Int -> Int -> Int
             logNat :: Bool -> Int -> Int -> Int

  Rules: log(x, y) -> logNat(x >= 0 /\ y >= 2, x, y) | x = x /\ y = y
         logNat(true, x, y) -> cond(x <= y, x, y) | x = x /\ y = y
         cond(true, x, y) -> 1 | x = x /\ y = y
         cond(false, x, y) -> 2 * log(x, y * y) | x = x /\ y = 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, i, c), where:

  P1. (1) log#(x, y) => logNat#(x >= 0 /\ y >= 2, x, y) | x = x /\ y = y
      (2) logNat#(true, x, y) => cond#(x <= y, x, y) | x = x /\ y = y
      (3) cond#(false, x, y) => log#(x, y * y) | x = x /\ y = y

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

We chain DPs according to the following mapping:


  cond#(false, x, y) => logNat#(x >= 0 /\ y * y >= 2, x, y * y) | x = x /\ y = y /\ (x = x /\ y * y = y * y)  is obtained by chaining  cond#(false, x, y) => log#(x, y * y) | x = x /\ y = y and log#(x', y') => logNat#(x' >= 0 /\ y' >= 2, x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

cond#(false, x, y) => log#(x, y * y) | x = x /\ y = y

log#(x, y) => logNat#(x >= 0 /\ y >= 2, x, y) | x = x /\ y = y


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

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

  P2. (1) logNat#(true, x, y) => cond#(x <= y, x, y) | x = x /\ y = y
      (2) cond#(false, x, y) => logNat#(x >= 0 /\ y * y >= 2, x, y * y) | x = x /\ y = y /\ (x = x /\ y * y = y * y)

***** We apply the Usable Rules Processor on D2 = (P2, R, i, c).

We obtain 0 usable rules (out of 4 rules in the input problem).

Processor output: { D3 = (P2, {}, i, c) }.

***** No progress could be made on DP problem D3 = (P2, {}, i, c).