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

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

  Rules: quot(0, s(X), s(Y)) -> 0
         quot(s(U), s(V), W) -> quot(U, V, W)
         quot(P, 0, s(X1)) -> s(quot(P, s(X1), s(X1)))
         map(Z1, nil) -> nil
         map(G1, cons(V1, W1)) -> cons(G1(V1), map(G1, W1))
         filter(J1, nil) -> nil
         filter(F2, cons(Y2, U2)) -> filter2(F2(Y2), F2, Y2, U2)
         filter2(true, H2, W2, P2) -> cons(W2, filter(H2, P2))
         filter2(false, F3, Y3, U3) -> filter(F3, U3)

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) quot#(s(U), s(V), W) => quot#(U, V, W)
      (2) quot#(P, 0, s(X1)) => quot#(P, s(X1), s(X1))
      (3) map#(G1, cons(V1, W1)) => map#(G1, W1)
      (4) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (5) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (6) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

We compute a graph approximation with the following edges:

  1: 1 2 
  2: 1 
  3: 3 
  4: 5 6 
  5: 4 
  6: 4 

There are 3 SCCs.

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

  P2. (1) quot#(s(U), s(V), W) => quot#(U, V, W)
      (2) quot#(P, 0, s(X1)) => quot#(P, s(X1), s(X1))

  P3. (1) map#(G1, cons(V1, W1)) => map#(G1, W1)

  P4. (1) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (2) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (3) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

We use the following projection function:

  nu(quot#) = 1

We thus have:

  (1) s(U) |>|  U
  (2) P    |>=| P

We may remove the strictly oriented DPs.

Processor output: { D5 = (P5, R, i, c) }, where:

  P5. (1) quot#(P, 0, s(X1)) => quot#(P, s(X1), s(X1))

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(V1, W1) |>| W1

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

Processor output: { }.

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

We use the following projection function:

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

We thus have:

  (1) cons(Y2, U2) |>|  U2
  (2) P2           |>=| P2
  (3) U3           |>=| U3

We may remove the strictly oriented DPs.

Processor output: { D6 = (P6, R, i, c) }, where:

  P6. (1) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (2) filter2#(false, F3, Y3, U3) => filter#(F3, U3)

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

We compute a graph approximation with the following edges:

  1:

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

Processor output: { }.

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

We compute a graph approximation with the following edges:

  1:
  2:

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

Processor output: { }.