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

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

  Rules: f(X, c(X), c(Y)) -> f(Y, Y, f(Y, X, Y))
         f(s(U), V, W) -> f(U, s(c(V)), c(W))
         f(c(P), P, X1) -> c(X1)
         g(Y1, U1) -> Y1
         g(V1, W1) -> W1
         map(J1, nil) -> nil
         map(F2, cons(Y2, U2)) -> cons(F2(Y2), map(F2, U2))
         filter(H2, nil) -> nil
         filter(I2, cons(P2, X3)) -> filter2(I2(P2), I2, P2, X3)
         filter2(true, Z3, U3, V3) -> cons(U3, filter(Z3, V3))
         filter2(false, I3, P3, X4) -> filter(I3, X4)

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#(X, c(X), c(Y)) => f#(Y, X, Y)
      (2) f#(X, c(X), c(Y)) => f#(Y, Y, f(Y, X, Y))
      (3) f#(s(U), V, W) => f#(U, s(c(V)), c(W))
      (4) map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (5) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (6) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (7) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

We compute a graph approximation with the following edges:

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

There are 4 SCCs.

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

  P2. (1) f#(s(U), V, W) => f#(U, s(c(V)), c(W))

  P3. (1) f#(X, c(X), c(Y)) => f#(Y, X, Y)
      (2) f#(X, c(X), c(Y)) => f#(Y, Y, f(Y, X, Y))

  P4. (1) map#(F2, cons(Y2, U2)) => map#(F2, U2)

  P5. (1) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (2) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (3) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** 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(U) |>| U

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

Processor output: { }.

***** No progress could be made on DP problem D3 = (P3, R, f, c).