We consider termination of the STRS with no additional rule schemes:

  Signature: 0      :: b
             cons   :: b -> a -> a
             curry  :: (b -> b -> b) -> b -> b -> b
             double :: a -> a
             inc    :: a -> a
             map    :: (b -> b) -> a -> a
             nil    :: a
             plus   :: b -> b -> b
             s      :: b -> b
             times  :: b -> b -> b

  Rules: plus(0, X) -> X
         plus(s(Y), U) -> s(plus(Y, U))
         times(0, V) -> 0
         times(s(W), P) -> plus(times(W, P), P)
         curry(F1, Y1, U1) -> F1(Y1, U1)
         map(H1, nil) -> nil
         map(I1, cons(P1, X2)) -> cons(I1(P1), map(I1, X2))
         inc -> map(curry(plus, s(0)))
         double -> map(curry(times, s(s(0))))

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)  plus#(s(Y), U) => plus#(Y, U)
      (2)  times#(s(W), P) => times#(W, P)
      (3)  times#(s(W), P) => plus#(times(W, P), P)
      (4)  map#(I1, cons(P1, X2)) => map#(I1, X2)
      (5)  inc#(arg1) => plus#(fresh1, fresh2)
      (6)  inc#(arg1) => curry#(plus, s(0), fresh1)
      (7)  inc#(arg1) => map#(curry(plus, s(0)), arg1)
      (8)  double#(arg1) => times#(fresh1, fresh2)
      (9)  double#(arg1) => curry#(times, s(s(0)), fresh1)
      (10) double#(arg1) => map#(curry(times, s(s(0))), arg1)

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

We compute a graph approximation with the following edges:

  1:  1 
  2:  2 3 
  3:  1 
  4:  4 
  5:  1 
  6:
  7:  4 
  8:  2 3 
  9:
  10: 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) plus#(s(Y), U) => plus#(Y, U)

  P3. (1) times#(s(W), P) => times#(W, P)

  P4. (1) map#(I1, cons(P1, X2)) => map#(I1, X2)

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

We use the following projection function:

  nu(plus#) = 1

We thus have:

  (1) s(Y) |>| Y

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

Processor output: { }.

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

We use the following projection function:

  nu(times#) = 1

We thus have:

  (1) s(W) |>| W

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

Processor output: { }.

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(P1, X2) |>| X2

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

Processor output: { }.