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

  Signature: eval :: Int -> Int -> Int -> Int -> Int -> o

  Rules: eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) | j < bn /\ i < an /\ cn = cn
         eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i + 1, j) | j < bn /\ i < an /\ cn = cn
         eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i + 1, j) | j >= bn /\ i < an /\ cn = cn
         eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) | j < bn /\ i >= an /\ cn = cn

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) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i, j + 1) | j < bn /\ i < an /\ cn = cn
      (2) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i + 1, j) | j < bn /\ i < an /\ cn = cn
      (3) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i + 1, j) | j >= bn /\ i < an /\ cn = cn
      (4) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i, j + 1) | j < bn /\ i >= an /\ cn = cn

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

We compute a graph approximation with the following edges:

  1: 1 2 3 
  2: 1 2 4 
  3: 3 
  4: 4 

There are 3 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) }, where:

  P2. (1) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i, j + 1) | j < bn /\ i >= an /\ cn = cn

  P3. (1) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i + 1, j) | j >= bn /\ i < an /\ cn = cn

  P4. (1) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i, j + 1) | j < bn /\ i < an /\ cn = cn
      (2) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i + 1, j) | j < bn /\ i < an /\ cn = cn

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

We use the following integer mapping:

  J(eval#) = arg_2 - arg_5

We thus have:

  (1) j < bn /\ i >= an /\ cn = cn |= bn - j > bn - (j + 1) (and bn - j >= 0)

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

Processor output: { }.

***** We apply the Integer Function Processor on D3 = (P3, R, i, c).

We use the following integer mapping:

  J(eval#) = arg_1 - arg_4

We thus have:

  (1) j >= bn /\ i < an /\ cn = cn |= an - i > an - (i + 1) (and an - i >= 0)

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

Processor output: { }.

***** We apply the Integer Function Processor on D4 = (P4, R, i, c).

We use the following integer mapping:

  J(eval#) = arg_1 - arg_4 - 1

We thus have:

  (1) j < bn /\ i < an /\ cn = cn |= an - i - 1 >= an - i - 1
  (2) j < bn /\ i < an /\ cn = cn |= an - i - 1 >  an - (i + 1) - 1 (and an - i - 1 >= 0)

We may remove the strictly oriented DPs, which yields:

Processor output: { D5 = (P5, R, i, c) }, where:

  P5. (1) eval#(an, bn, cn, i, j) => eval#(an, bn, cn + 1, i, j + 1) | j < bn /\ i < an /\ cn = cn

***** We apply the Integer Function Processor on D5 = (P5, R, i, c).

We use the following integer mapping:

  J(eval#) = arg_2 - arg_5

We thus have:

  (1) j < bn /\ i < an /\ cn = cn |= bn - j > bn - (j + 1) (and bn - j >= 0)

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

Processor output: { }.