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

  Signature: append  :: a -> a -> a
             combine :: a -> a -> a
             cons    :: a -> a -> a
             levels  :: a -> a
             map     :: (a -> a) -> a -> a
             nil     :: a
             node    :: a -> a -> a
             zip     :: a -> a -> a

  Rules: map(F, nil) -> nil
         map(Z, cons(U, V)) -> cons(Z(U), map(Z, V))
         append(W, nil) -> W
         append(nil, P) -> P
         append(cons(X1, Y1), U1) -> cons(X1, append(Y1, U1))
         zip(nil, V1) -> V1
         zip(W1, nil) -> W1
         zip(cons(P1, X2), cons(Y2, U2)) -> cons(append(P1, Y2), zip(X2, U2))
         combine(V2, nil) -> V2
         combine(W2, cons(P2, X3)) -> combine(zip(W2, P2), X3)
         levels(node(Y3, U3)) -> cons(cons(Y3, nil), combine(nil, map(levels, 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) map#(Z, cons(U, V)) => map#(Z, V)
      (2) append#(cons(X1, Y1), U1) => append#(Y1, U1)
      (3) zip#(cons(P1, X2), cons(Y2, U2)) => append#(P1, Y2)
      (4) zip#(cons(P1, X2), cons(Y2, U2)) => zip#(X2, U2)
      (5) combine#(W2, cons(P2, X3)) => zip#(W2, P2)
      (6) combine#(W2, cons(P2, X3)) => combine#(zip(W2, P2), X3)
      (7) levels#(node(Y3, U3)) => levels#(fresh1)
      (8) levels#(node(Y3, U3)) => map#(levels, U3)
      (9) levels#(node(Y3, U3)) => combine#(nil, map(levels, 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 
  3: 2 
  4: 3 4 
  5: 3 4 
  6: 5 6 
  7: 7 8 9 
  8: 1 
  9: 5 6 

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) map#(Z, cons(U, V)) => map#(Z, V)

  P3. (1) append#(cons(X1, Y1), U1) => append#(Y1, U1)

  P4. (1) zip#(cons(P1, X2), cons(Y2, U2)) => zip#(X2, U2)

  P5. (1) combine#(W2, cons(P2, X3)) => combine#(zip(W2, P2), X3)

  P6. (1) levels#(node(Y3, U3)) => levels#(fresh1)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(U, V) |>| V

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

Processor output: { }.

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

We use the following projection function:

  nu(append#) = 1

We thus have:

  (1) cons(X1, Y1) |>| Y1

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(zip#) = 1

We thus have:

  (1) cons(P1, X2) |>| X2

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(combine#) = 2

We thus have:

  (1) cons(P2, X3) |>| X3

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

Processor output: { }.

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

We obtain 0 usable rules (out of 11 rules in the input problem).

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

***** No progress could be made on DP problem D7 = (P6, {}, i, c).