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)  times#(X, plus(Y, s(U))) => times#(s(U), 0)
      (2)  times#(X, plus(Y, s(U))) => plus#(Y, times(s(U), 0))
      (3)  times#(X, plus(Y, s(U))) => times#(X, plus(Y, times(s(U), 0)))
      (4)  times#(X, plus(Y, s(U))) => times#(X, s(U))
      (5)  times#(X, plus(Y, s(U))) => plus#(times(X, plus(Y, times(s(U), 0))), times(X, s(U)))
      (6)  times#(W, s(P)) => times#(W, P)
      (7)  times#(W, s(P)) => plus#(times(W, P), W)
      (8)  plus#(Y1, s(U1)) => plus#(Y1, U1)
      (9)  map#(I1, cons(P1, X2)) => map#(I1, X2)
      (10) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (11) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (12) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

Considering the 4 SCCs, this DP problem is split into the following new problems.

  P2. (1) plus#(Y1, s(U1)) => plus#(Y1, U1)

  P3. (1) times#(X, plus(Y, s(U))) => times#(X, plus(Y, times(s(U), 0)))
      (2) times#(X, plus(Y, s(U))) => times#(X, s(U))
      (3) times#(W, s(P)) => times#(W, P)

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

  P5. (1) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (2) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (3) filter2#(false, G3, V3, W3) => filter#(G3, W3)

***** We apply the Subterm Criterion Processor on P2.

We use the following projection function:

  nu(plus#) = 2

We thus have:

  (1) s(U1) |>| U1

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

***** No progress could be made on DP problem P3.