We consider termination of the STRS with no additional rule schemes:

  Signature: 0      :: a
             cons   :: b -> c -> c
             d      :: a -> a -> c
             false  :: c
             filter :: (b -> c) -> c -> c
             gtr    :: a -> a -> c
             if     :: c -> c -> c -> c
             len    :: c -> a
             nil    :: c
             s      :: a -> a
             sub    :: a -> a -> a
             true   :: c

  Rules: 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))

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R, i, c), where:

  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 D1 = (P1, R, i, c).

We compute a graph approximation with the following edges:

  1:  1 
  2:  2 
  3:  2 
  4:  1 
  5:  3 4 5 6 
  6:
  7:  7 
  8:  8 9 10 
  9:  8 9 10 
  10:

There are 5 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) ; D5 = (P5, R, i, c) ; D6 = (P6, R, i, c) }, where:

  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 D2 = (P2, R, i, c).

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.

Processor output: { }.

***** We apply the Subterm Criterion Processor on D3 = (P3, R, i, c).

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.

Processor output: { }.

***** We apply the Usable Rules Processor on D4 = (P4, R, i, c).

We obtain 2 usable rules (out of 13 rules in the input problem).

Processor output: { D7 = (P4, R2, i, c) }, where:

  R2. (1) sub(W, 0) -> W
      (2) sub(s(P), s(X1)) -> sub(P, X1)

***** We apply the Subterm Criterion Processor on D5 = (P5, R, i, c).

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.

Processor output: { }.

***** We apply the Subterm Criterion Processor on D6 = (P6, R, i, c).

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.

Processor output: { }.

***** We apply the Usable rules with respect to HORPO Processor on D7 = (P4, R2, i, c).

Constrained HORPO yields:

  d#2(s1(X2), s1(Y2)) (>) d#2(s1(X2), sub2(Y2, X2))
  sub2(W, 00) (>=) W
  sub2(s1(P), s1(X1)) (>=) sub2(P, X1)

We do this using the following settings:

* Disregarded arguments:

  sub2 2 

* Precedence and permutation:

    s1    { 1 }
  > 00    { }
  = d#2   { 2 } 1
  = sub2  { 1 }

* Well-founded theory orderings:

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

Processor output: { }.