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
             inc   :: a -> a
             map   :: (b -> b) -> a -> a
             nil   :: a
             plus  :: b -> b -> b
             s     :: b -> b

  Rules: plus(0, X) -> X
         plus(s(Y), U) -> s(plus(Y, U))
         map(H, nil) -> nil
         map(I, cons(P, X1)) -> cons(I(P), map(I, X1))
         curry(Z1, U1, V1) -> Z1(U1, V1)
         inc -> map(curry(plus, 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) map#(I, cons(P, X1)) => map#(I, X1)
      (3) inc#(arg1) => plus#(fresh1, fresh2)
      (4) inc#(arg1) => curry#(plus, s(0), fresh1)
      (5) inc#(arg1) => map#(curry(plus, 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: 1 
  4:
  5: 2 

There are 2 SCCs.

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

  P2. (1) plus#(s(Y), U) => plus#(Y, U)

  P3. (1) map#(I, cons(P, X1)) => map#(I, X1)

***** 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(map#) = 2

We thus have:

  (1) cons(P, X1) |>| X1

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

Processor output: { }.