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
             log     :: a -> a
             map     :: (c -> c) -> d -> d
             minus   :: a -> a -> a
             nil     :: d
             quot    :: a -> a -> a
             s       :: a -> a
             true    :: b

  Rules: minus(X, 0) -> X
         minus(s(Y), s(U)) -> minus(Y, U)
         quot(0, s(V)) -> 0
         quot(s(W), s(P)) -> s(quot(minus(W, P), s(P)))
         log(s(0)) -> 0
         log(s(s(X1))) -> s(log(s(quot(X1, s(s(0))))))
         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) minus#(s(Y), s(U)) => minus#(Y, U)
      (2) quot#(s(W), s(P)) => minus#(W, P)
      (3) quot#(s(W), s(P)) => quot#(minus(W, P), s(P))
      (4) log#(s(s(X1))) => quot#(X1, s(s(0)))
      (5) log#(s(s(X1))) => log#(s(quot(X1, s(s(0)))))
      (6) map#(G1, cons(V1, W1)) => map#(G1, W1)
      (7) filter#(F2, cons(Y2, U2)) => filter2#(F2(Y2), F2, Y2, U2)
      (8) filter2#(true, H2, W2, P2) => filter#(H2, P2)
      (9) 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: 1 
  3: 2 3 
  4: 2 3 
  5: 4 5 
  6: 6 
  7: 8 9 
  8: 7 
  9: 7 

There are 5 SCCs.

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

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

  P3. (1) quot#(s(W), s(P)) => quot#(minus(W, P), s(P))

  P4. (1) log#(s(s(X1))) => log#(s(quot(X1, s(s(0)))))

  P5. (1) map#(G1, cons(V1, W1)) => map#(G1, W1)

  P6. (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(minus#) = 1

We thus have:

  (1) s(Y) |>| Y

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, i, c).