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

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

  Rules: plus(X, 0) -> X
         plus(Y, s(U)) -> s(plus(Y, U))
         f(0, s(0), V) -> f(V, plus(V, V), V)
         g(W, P) -> W
         g(X1, Y1) -> Y1
         map(G1, nil) -> nil
         map(H1, cons(W1, P1)) -> cons(H1(W1), map(H1, P1))
         filter(F2, nil) -> nil
         filter(Z2, cons(U2, V2)) -> filter2(Z2(U2), Z2, U2, V2)
         filter2(true, I2, P2, X3) -> cons(P2, filter(I2, X3))
         filter2(false, Z3, U3, V3) -> filter(Z3, V3)

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) plus#(Y, s(U)) => plus#(Y, U)
      (2) f#(0, s(0), V) => plus#(V, V)
      (3) f#(0, s(0), V) => f#(V, plus(V, V), V)
      (4) map#(H1, cons(W1, P1)) => map#(H1, P1)
      (5) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (6) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (7) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We compute a graph approximation with the following edges:

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

There are 4 SCCs.

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

  P2. (1) plus#(Y, s(U)) => plus#(Y, U)

  P3. (1) f#(0, s(0), V) => f#(V, plus(V, V), V)

  P4. (1) map#(H1, cons(W1, P1)) => map#(H1, P1)

  P5. (1) filter#(Z2, cons(U2, V2)) => filter2#(Z2(U2), Z2, U2, V2)
      (2) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (3) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

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

We use the following projection function:

  nu(plus#) = 2

We thus have:

  (1) s(U) |>| U

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

Processor output: { }.

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

We obtain 2 usable rules (out of 11 rules in the input problem).

Processor output: { D6 = (P3, R2, i, c) }, where:

  R2. (1) plus(X, 0) -> X
      (2) plus(Y, s(U)) -> s(plus(Y, U))

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(W1, P1) |>| P1

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

Processor output: { }.

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

We use the following projection function:

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

We thus have:

  (1) cons(U2, V2) |>|  V2
  (2) X3           |>=| X3
  (3) V3           |>=| V3

We may remove the strictly oriented DPs.

Processor output: { D7 = (P6, R, i, c) }, where:

  P6. (1) filter2#(true, I2, P2, X3) => filter#(I2, X3)
      (2) filter2#(false, Z3, U3, V3) => filter#(Z3, V3)

***** No progress could be made on DP problem D6 = (P3, R2, i, c).