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)  sub#(s(P), s(X1)) => sub#(P, X1)
      (2)  gtr#(s(V1), s(W1)) => gtr#(V1, W1)
      (3)  d#(s(X2), s(Y2)) => gtr#(X2, Y2)
      (4)  d#(s(X2), s(Y2)) => sub#(Y2, X2)
      (5)  d#(s(X2), s(Y2)) => d#(s(X2), sub(Y2, X2))
      (6)  d#(s(X2), s(Y2)) => if#(gtr(X2, Y2), false, d(s(X2), sub(Y2, X2)))
      (7)  len#(cons(U2, V2)) => len#(V2)
      (8)  filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)
      (9)  filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)
      (10) filter#(J2, cons(X3, Y3)) => if#(J2(X3), cons(X3, filter(J2, Y3)), filter(J2, Y3))

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

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

  P2. (1) sub#(s(P), s(X1)) => sub#(P, X1)

  P3. (1) gtr#(s(V1), s(W1)) => gtr#(V1, W1)

  P4. (1) d#(s(X2), s(Y2)) => d#(s(X2), sub(Y2, X2))

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

  P6. (1) filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)
      (2) filter#(J2, cons(X3, Y3)) => filter#(J2, Y3)

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

We use the following projection function:

  nu(sub#) = 1

We thus have:

  (1) s(P) |>| P

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(gtr#) = 1

We thus have:

  (1) s(V1) |>| V1

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

***** We apply the horpo Processor on P4.

Constrained HORPO yields:

  d#(s(X2), s(Y2)) (>) d#(s(X2), sub(Y2, X2))
  if(true, X, Y) (>=) X
  if(false, U, V) (>=) V
  sub(W, 0) (>=) W
  sub(s(P), s(X1)) (>=) sub(P, X1)
  gtr(0, Y1) (>=) false
  gtr(s(U1), 0) (>=) true
  gtr(s(V1), s(W1)) (>=) gtr(V1, W1)
  d(P1, 0) (>=) true
  d(s(X2), s(Y2)) (>=) if(gtr(X2, Y2), false, d(s(X2), sub(Y2, X2)))
  len(nil) (>=) 0
  len(cons(U2, V2)) (>=) s(len(V2))
  filter(I2, nil) (>=) nil
  filter(J2, cons(X3, Y3)) (>=) if(J2(X3), cons(X3, filter(J2, Y3)), filter(J2, Y3))

We do this using the following settings:

* Precedence and status (for non-mentioned symbols the precedence is irrelevant and the status is Lex):

  len               (status: Lex)   >
  filter            (status: Mul_2) >
  cons = nil        (status: Lex)   >
  0                 (status: Lex)   >
  gtr               (status: Mul_2) >
  s                 (status: Lex)   >
  false = if = sub  (status: Lex)   >
  d                 (status: Lex)   >
  d#                (status: Mul_2) >
  true              (status: Lex)

* Well-founded theory orderings:

  [>]_{Bool} = {(true,false)}
  [>]_{Int}  = {(x,y) | x < 1000 /\ x < y }

* Filter:

  d      disregards argument(s) 1 
  d#     disregards argument(s) 1 
  filter disregards argument(s) 1 
  gtr    disregards argument(s) 1 
  if     disregards argument(s) 1 
  sub    disregards argument(s) 2 

All dependency pairs were removed.

***** We apply the Subterm Criterion Processor on P5.

We use the following projection function:

  nu(len#) = 1

We thus have:

  (1) cons(U2, V2) |>| V2

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

***** We apply the Subterm Criterion Processor on P6.

We use the following projection function:

  nu(filter#) = 2

We thus have:

  (1) cons(X3, Y3) |>| Y3
  (2) cons(X3, Y3) |>| Y3

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