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

  Signature: 0       :: d
             cons    :: d -> d -> d
             f       :: d -> a
             false   :: c
             filter  :: (d -> c) -> d -> d
             filter2 :: c -> (d -> c) -> d -> d -> d
             g       :: d -> d
             h       :: d -> b
             map     :: (d -> d) -> d -> d
             nil     :: d
             s       :: d -> d
             true    :: c

  Rules: f(s(X)) -> f(X)
         g(cons(0, Y)) -> g(Y)
         g(cons(s(U), V)) -> s(U)
         h(cons(W, P)) -> h(g(cons(W, P)))
         map(F1, nil) -> nil
         map(Z1, cons(U1, V1)) -> cons(Z1(U1), map(Z1, V1))
         filter(I1, nil) -> nil
         filter(J1, cons(X2, Y2)) -> filter2(J1(X2), J1, X2, Y2)
         filter2(true, G2, V2, W2) -> cons(V2, filter(G2, W2))
         filter2(false, J2, X3, 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, f, c), where:

  P1. (1) f#(s(X)) => f#(X)
      (2) g#(cons(0, Y)) => g#(Y)
      (3) h#(cons(W, P)) => g#(cons(W, P))
      (4) h#(cons(W, P)) => h#(g(cons(W, P)))
      (5) map#(Z1, cons(U1, V1)) => map#(Z1, V1)
      (6) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (7) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (8) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

***** We apply the Graph Processor on D1 = (P1, R, f, c).

We compute a graph approximation with the following edges:

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

There are 5 SCCs.

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

  P2. (1) f#(s(X)) => f#(X)

  P3. (1) g#(cons(0, Y)) => g#(Y)

  P4. (1) h#(cons(W, P)) => h#(g(cons(W, P)))

  P5. (1) map#(Z1, cons(U1, V1)) => map#(Z1, V1)

  P6. (1) filter#(J1, cons(X2, Y2)) => filter2#(J1(X2), J1, X2, Y2)
      (2) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (3) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

***** We apply the Subterm Criterion Processor on D2 = (P2, R, f, c).

We use the following projection function:

  nu(f#) = 1

We thus have:

  (1) s(X) |>| X

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, f, c).

We use the following projection function:

  nu(g#) = 1

We thus have:

  (1) cons(0, Y) |>| Y

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

Processor output: { }.

***** We apply the Reduction Pair [with HORPO] Processor on D4 = (P4, R, f, c).

Constrained HORPO yields:

  h#(cons(W, P)) (>) h#(g(cons(W, P)))
  f(s(X)) (>=) f(X)
  g(cons(0, Y)) (>=) g(Y)
  g(cons(s(U), V)) (>=) s(U)
  h(cons(W, P)) (>=) h(g(cons(W, P)))
  map(F1, nil) (>=) nil
  map(Z1, cons(U1, V1)) (>=) cons(Z1(U1), map(Z1, V1))
  filter(I1, nil) (>=) nil
  filter(J1, cons(X2, Y2)) (>=) filter2(J1(X2), J1, X2, Y2)
  filter2(true, G2, V2, W2) (>=) cons(V2, filter(G2, W2))
  filter2(false, J2, X3, Y3) (>=) filter(J2, Y3)

We do this using the following settings:

* Disregarded arguments:

  f 1 
  g 1 
  h 1 
  s 1 

* Precedence and permutation:

    false    { }
  > 0        { }
  = h#       { 1 }
  > map      { 1 2 }
  = nil      { }
  > filter2  { 2 3 4 } _ _ 1
  = filter   { 1 2 }
  = true     { }
  > cons     { 1 2 }
  > h        { }
  > f        { }
  = g        { }
  = s        { }

* Well-founded theory orderings:

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

All dependency pairs were removed.

Processor output: { }.

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U1, V1) |>| V1

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, f, c).

We use the following projection function:

  nu(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(X2, Y2) |>|  Y2
  (2) W2           |>=| W2
  (3) Y3           |>=| Y3

We may remove the strictly oriented DPs.

Processor output: { D7 = (P7, R, f, c) }, where:

  P7. (1) filter2#(true, G2, V2, W2) => filter#(G2, W2)
      (2) filter2#(false, J2, X3, Y3) => filter#(J2, Y3)

***** We apply the Graph Processor on D7 = (P7, R, f, c).

We compute a graph approximation with the following edges:

  1:
  2:

As there are no SCCs, this DP problem is removed.

Processor output: { }.