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)  eq#(s(V), s(U)) => eq#(V, U)
      (2)  le#(s(Y1), s(X1)) => le#(Y1, X1)
      (3)  min#(cons(W1, cons(V1, P1))) => le#(W1, V1)
      (4)  min#(cons(W1, cons(V1, P1))) => if_min#(le(W1, V1), cons(W1, cons(V1, P1)))
      (5)  if_min#(true, cons(Y2, cons(X2, U2))) => min#(cons(Y2, U2))
      (6)  if_min#(false, cons(W2, cons(V2, P2))) => min#(cons(V2, P2))
      (7)  replace#(W3, V3, cons(U3, P3)) => eq#(W3, U3)
      (8)  replace#(W3, V3, cons(U3, P3)) => if_replace#(eq(W3, U3), W3, V3, cons(U3, P3))
      (9)  if_replace#(false, X5, P4, cons(W4, Y5)) => replace#(X5, P4, Y5)
      (10) sort#(cons(U5, V5)) => min#(cons(U5, V5))
      (11) sort#(cons(U5, V5)) => min#(cons(U5, V5))
      (12) sort#(cons(U5, V5)) => replace#(min(cons(U5, V5)), U5, V5)
      (13) sort#(cons(U5, V5)) => sort#(replace(min(cons(U5, V5)), U5, V5))
      (14) map#(J5, cons(X6, Y6)) => map#(J5, Y6)
      (15) filter#(H6, cons(W6, P6)) => filter2#(H6(W6), H6, W6, P6)
      (16) filter2#(true, F7, Y7, U7) => filter#(F7, U7)
      (17) filter2#(false, H7, W7, P7) => filter#(H7, P7)

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

Considering the 7 SCCs, this DP problem is split into the following new problems.

  P2. (1) eq#(s(V), s(U)) => eq#(V, U)

  P3. (1) le#(s(Y1), s(X1)) => le#(Y1, X1)

  P4. (1) min#(cons(W1, cons(V1, P1))) => if_min#(le(W1, V1), cons(W1, cons(V1, P1)))
      (2) if_min#(true, cons(Y2, cons(X2, U2))) => min#(cons(Y2, U2))
      (3) if_min#(false, cons(W2, cons(V2, P2))) => min#(cons(V2, P2))

  P5. (1) replace#(W3, V3, cons(U3, P3)) => if_replace#(eq(W3, U3), W3, V3, cons(U3, P3))
      (2) if_replace#(false, X5, P4, cons(W4, Y5)) => replace#(X5, P4, Y5)

  P6. (1) sort#(cons(U5, V5)) => sort#(replace(min(cons(U5, V5)), U5, V5))

  P7. (1) map#(J5, cons(X6, Y6)) => map#(J5, Y6)

  P8. (1) filter#(H6, cons(W6, P6)) => filter2#(H6(W6), H6, W6, P6)
      (2) filter2#(true, F7, Y7, U7) => filter#(F7, U7)
      (3) filter2#(false, H7, W7, P7) => filter#(H7, P7)

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

We use the following projection function:

  nu(eq#) = 1

We thus have:

  (1) s(V) |>| V

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(le#) = 1

We thus have:

  (1) s(Y1) |>| Y1

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

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