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)  minus#(minus(V, W), P) => plus#(W, P)
      (3)  minus#(minus(V, W), P) => minus#(V, plus(W, P))
      (4)  quot#(s(Y1), s(U1)) => minus#(Y1, U1)
      (5)  quot#(s(Y1), s(U1)) => quot#(minus(Y1, U1), s(U1))
      (6)  plus#(s(W1), P1) => plus#(W1, P1)
      (7)  app#(cons(W2, V2), U2) => app#(V2, U2)
      (8)  sum#(cons(Y3, cons(U3, X3))) => plus#(Y3, U3)
      (9)  sum#(cons(Y3, cons(U3, X3))) => sum#(cons(plus(Y3, U3), X3))
      (10) sum#(app(W3, cons(P3, cons(X4, V3)))) => sum#(cons(P3, cons(X4, V3)))
      (11) sum#(app(W3, cons(P3, cons(X4, V3)))) => app#(W3, sum(cons(P3, cons(X4, V3))))
      (12) sum#(app(W3, cons(P3, cons(X4, V3)))) => sum#(app(W3, sum(cons(P3, cons(X4, V3)))))
      (13) map#(G4, cons(V4, W4)) => map#(G4, W4)
      (14) filter#(F5, cons(Y5, U5)) => filter2#(F5(Y5), F5, Y5, U5)
      (15) filter2#(true, H5, W5, P5) => filter#(H5, P5)
      (16) filter2#(false, F6, Y6, U6) => filter#(F6, U6)

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

Considering the 8 SCCs, this DP problem is split into the following new problems.

  P2. (1) plus#(s(W1), P1) => plus#(W1, P1)

  P3. (1) minus#(s(Y), s(U)) => minus#(Y, U)
      (2) minus#(minus(V, W), P) => minus#(V, plus(W, P))

  P4. (1) quot#(s(Y1), s(U1)) => quot#(minus(Y1, U1), s(U1))

  P5. (1) app#(cons(W2, V2), U2) => app#(V2, U2)

  P6. (1) sum#(cons(Y3, cons(U3, X3))) => sum#(cons(plus(Y3, U3), X3))

  P7. (1) sum#(app(W3, cons(P3, cons(X4, V3)))) => sum#(app(W3, sum(cons(P3, cons(X4, V3)))))

  P8. (1) map#(G4, cons(V4, W4)) => map#(G4, W4)

  P9. (1) filter#(F5, cons(Y5, U5)) => filter2#(F5(Y5), F5, Y5, U5)
      (2) filter2#(true, H5, W5, P5) => filter#(H5, P5)
      (3) filter2#(false, F6, Y6, U6) => filter#(F6, U6)

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

We use the following projection function:

  nu(plus#) = 1

We thus have:

  (1) s(W1) |>| W1

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(minus#) = 1

We thus have:

  (1) s(Y)        |>| Y
  (2) minus(V, W) |>| V

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

***** No progress could be made on DP problem P4.