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

  Signature: 0       :: a
             ack     :: a -> a -> 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
             succ    :: a -> a
             true    :: b

  Rules: ack(0, X) -> succ(X)
         ack(succ(Y), U) -> ack(Y, succ(0))
         ack(succ(V), succ(W)) -> ack(V, ack(succ(V), W))
         map(J, nil) -> nil
         map(F1, cons(Y1, U1)) -> cons(F1(Y1), map(F1, U1))
         filter(H1, nil) -> nil
         filter(I1, cons(P1, X2)) -> filter2(I1(P1), I1, P1, X2)
         filter2(true, Z2, U2, V2) -> cons(U2, filter(Z2, V2))
         filter2(false, I2, P2, X3) -> filter(I2, X3)

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) ack#(succ(Y), U) => ack#(Y, succ(0))
      (2) ack#(succ(V), succ(W)) => ack#(succ(V), W)
      (3) ack#(succ(V), succ(W)) => ack#(V, ack(succ(V), W))
      (4) map#(F1, cons(Y1, U1)) => map#(F1, U1)
      (5) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (6) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (7) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We compute a graph approximation with the following edges:

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

There are 3 SCCs.

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

  P2. (1) ack#(succ(Y), U) => ack#(Y, succ(0))
      (2) ack#(succ(V), succ(W)) => ack#(succ(V), W)
      (3) ack#(succ(V), succ(W)) => ack#(V, ack(succ(V), W))

  P3. (1) map#(F1, cons(Y1, U1)) => map#(F1, U1)

  P4. (1) filter#(I1, cons(P1, X2)) => filter2#(I1(P1), I1, P1, X2)
      (2) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (3) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We use the following projection function:

  nu(ack#) = 1

We thus have:

  (1) succ(Y) |>|  Y
  (2) succ(V) |>=| succ(V)
  (3) succ(V) |>|  V

We may remove the strictly oriented DPs.

Processor output: { D5 = (P5, R, i, c) }, where:

  P5. (1) ack#(succ(V), succ(W)) => ack#(succ(V), W)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(Y1, U1) |>| U1

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

We thus have:

  (1) cons(P1, X2) |>|  X2
  (2) V2           |>=| V2
  (3) X3           |>=| X3

We may remove the strictly oriented DPs.

Processor output: { D6 = (P6, R, i, c) }, where:

  P6. (1) filter2#(true, Z2, U2, V2) => filter#(Z2, V2)
      (2) filter2#(false, I2, P2, X3) => filter#(I2, X3)

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

We use the following projection function:

  nu(ack#) = 2

We thus have:

  (1) succ(W) |>| W

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

Processor output: { }.

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

We compute a graph approximation with the following edges:

  1:
  2:

As there are no SCCs, this DP problem is removed.

Processor output: { }.