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)  minus#(s(Y), s(U)) => minus#(Y, U)
      (2)  quot#(s(W), s(P)) => minus#(W, P)
      (3)  quot#(s(W), s(P)) => quot#(minus(W, P), s(P))
      (4)  plus#(s(Y1), U1) => plus#(Y1, U1)
      (5)  plus#(minus(V1, s(0)), minus(W1, s(s(P1)))) => minus#(W1, s(s(P1)))
      (6)  plus#(minus(V1, s(0)), minus(W1, s(s(P1)))) => minus#(V1, s(0))
      (7)  plus#(minus(V1, s(0)), minus(W1, s(s(P1)))) => plus#(minus(W1, s(s(P1))), minus(V1, s(0)))
      (8)  plus#(plus(X2, s(0)), plus(Y2, s(s(U2)))) => plus#(Y2, s(s(U2)))
      (9)  plus#(plus(X2, s(0)), plus(Y2, s(s(U2)))) => plus#(X2, s(0))
      (10) plus#(plus(X2, s(0)), plus(Y2, s(s(U2)))) => plus#(plus(Y2, s(s(U2))), plus(X2, s(0)))
      (11) map#(I2, cons(P2, X3)) => map#(I2, X3)
      (12) filter#(G3, cons(V3, W3)) => filter2#(G3(V3), G3, V3, W3)
      (13) filter2#(true, J3, X4, Y4) => filter#(J3, Y4)
      (14) filter2#(false, G4, V4, W4) => filter#(G4, W4)

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

Considering the 5 SCCs, this DP problem is split into the following new problems.

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

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

  P4. (1) plus#(s(Y1), U1) => plus#(Y1, U1)
      (2) plus#(minus(V1, s(0)), minus(W1, s(s(P1)))) => plus#(minus(W1, s(s(P1))), minus(V1, s(0)))
      (3) plus#(plus(X2, s(0)), plus(Y2, s(s(U2)))) => plus#(Y2, s(s(U2)))
      (4) plus#(plus(X2, s(0)), plus(Y2, s(s(U2)))) => plus#(X2, s(0))
      (5) plus#(plus(X2, s(0)), plus(Y2, s(s(U2)))) => plus#(plus(Y2, s(s(U2))), plus(X2, s(0)))

  P5. (1) map#(I2, cons(P2, X3)) => map#(I2, X3)

  P6. (1) filter#(G3, cons(V3, W3)) => filter2#(G3(V3), G3, V3, W3)
      (2) filter2#(true, J3, X4, Y4) => filter#(J3, Y4)
      (3) filter2#(false, G4, V4, W4) => filter#(G4, W4)

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

We use the following projection function:

  nu(minus#) = 1

We thus have:

  (1) s(Y) |>| Y

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

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