The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the following initial DP problem:

  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#(X{13}) => plus#(X{14}, X{15})
      (6)  inc#(X{13}) => curry#(plus, s(0), X{16})
      (7)  inc#(X{13}) => map#(curry(plus, s(0)), X{13})
      (8)  double#(X{17}) => times#(X{18}, X{19})
      (9)  double#(X{17}) => curry#(times, s(s(0)), X{20})
      (10) double#(X{17}) => map#(curry(times, s(s(0))), X{17})

***** We apply the Graph Processor on P1.

Considering the 3 SCCs, this DP problem is split into the following new problems.

  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 P2.

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.

***** We apply the Subterm Criterion Processor on P3.

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.

***** We apply the Subterm Criterion Processor on P4.

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.