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
             plus    :: a -> a -> a
             s       :: a -> a
             times   :: a -> a -> a
             true    :: b

  Rules: times(X, plus(Y, s(U))) -> plus(times(X, plus(Y, times(s(U), 0))), times(X, s(U)))
         times(V, 0) -> 0
         times(W, s(P)) -> plus(times(W, P), W)
         plus(X1, 0) -> X1
         plus(Y1, s(U1)) -> s(plus(Y1, U1))
         map(H1, nil) -> nil
         map(I1, cons(P1, X2)) -> cons(I1(P1), map(I1, X2))
         filter(Z2, nil) -> nil
         filter(G2, cons(V2, W2)) -> filter2(G2(V2), G2, V2, W2)
         filter2(true, J2, X3, Y3) -> cons(X3, filter(J2, Y3))
         filter2(false, G3, V3, W3) -> filter(G3, W3)

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)  times#(X, plus(Y, s(U))) => times#(s(U), 0)
      (2)  times#(X, plus(Y, s(U))) => plus#(Y, times(s(U), 0))
      (3)  times#(X, plus(Y, s(U))) => times#(X, plus(Y, times(s(U), 0)))
      (4)  times#(X, plus(Y, s(U))) => times#(X, s(U))
      (5)  times#(X, plus(Y, s(U))) => plus#(times(X, plus(Y, times(s(U), 0))), times(X, s(U)))
      (6)  times#(W, s(P)) => times#(W, P)
      (7)  times#(W, s(P)) => plus#(times(W, P), W)
      (8)  plus#(Y1, s(U1)) => plus#(Y1, U1)
      (9)  map#(I1, cons(P1, X2)) => map#(I1, X2)
      (10) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (11) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (12) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

We compute a graph approximation with the following edges:

  1:
  2:  8 
  3:  1 2 3 4 5 6 7 
  4:  6 7 
  5:  8 
  6:  1 2 3 4 5 6 7 
  7:  8 
  8:  8 
  9:  9 
  10: 11 12 
  11: 10 
  12: 10 

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) plus#(Y1, s(U1)) => plus#(Y1, U1)

  P3. (1) times#(X, plus(Y, s(U))) => times#(X, plus(Y, times(s(U), 0)))
      (2) times#(X, plus(Y, s(U))) => times#(X, s(U))
      (3) times#(W, s(P)) => times#(W, P)

  P4. (1) map#(I1, cons(P1, X2)) => map#(I1, X2)

  P5. (1) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (2) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (3) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

We use the following projection function:

  nu(plus#) = 2

We thus have:

  (1) s(U1) |>| U1

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).