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

  Signature: 0      :: b
             cons   :: b -> a -> a
             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)
         map(F1, nil) -> nil
         map(Z1, cons(U1, V1)) -> cons(Z1(U1), map(Z1, V1))
         inc -> map(plus(s(0)))
         double -> map(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#(Z1, cons(U1, V1)) => map#(Z1, V1)
      (5) inc#(arg1) => plus#(s(0), fresh1)
      (6) inc#(arg1) => map#(plus(s(0)), arg1)
      (7) double#(arg1) => times#(s(s(0)), fresh1)
      (8) double#(arg1) => map#(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: 4 
  7: 2 3 
  8: 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#(Z1, cons(U1, V1)) => map#(Z1, V1)

***** 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(U1, V1) |>| V1

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

Processor output: { }.