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

  Signature: 0       :: a
             cons    :: c -> d -> d
             double  :: a -> a
             false   :: b
             filter  :: (c -> b) -> d -> d
             filter2 :: b -> (c -> b) -> c -> d -> d
             map     :: (c -> c) -> d -> d
             minus   :: a -> a -> a
             nil     :: d
             plus    :: a -> a -> a
             s       :: a -> a
             true    :: b

  Rules: minus(X, 0) -> X
         minus(s(Y), s(U)) -> minus(Y, U)
         double(0) -> 0
         double(s(V)) -> s(s(double(V)))
         plus(0, W) -> W
         plus(s(P), X1) -> s(plus(P, X1))
         plus(s(Y1), U1) -> plus(Y1, s(U1))
         plus(s(V1), W1) -> s(plus(minus(V1, W1), double(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, i, c), where:

  P1. (1)  minus#(s(Y), s(U)) => minus#(Y, U)
      (2)  double#(s(V)) => double#(V)
      (3)  plus#(s(P), X1) => plus#(P, X1)
      (4)  plus#(s(Y1), U1) => plus#(Y1, s(U1))
      (5)  plus#(s(V1), W1) => minus#(V1, W1)
      (6)  plus#(s(V1), W1) => double#(W1)
      (7)  plus#(s(V1), W1) => plus#(minus(V1, W1), double(W1))
      (8)  map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (9)  filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (10) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (11) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

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) double#(s(V)) => double#(V)

  P4. (1) plus#(s(P), X1) => plus#(P, X1)
      (2) plus#(s(Y1), U1) => plus#(Y1, s(U1))
      (3) plus#(s(V1), W1) => plus#(minus(V1, W1), double(W1))

  P5. (1) map#(F2, cons(Y2, U2)) => map#(F2, U2)

  P6. (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, 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: { }.

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

We use the following projection function:

  nu(double#) = 1

We thus have:

  (1) s(V) |>| V

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

Processor output: { }.

***** We apply the Usable Rules Processor on D4 = (P4, R, i, c).

We obtain 4 usable rules (out of 14 rules in the input problem).

Processor output: { D7 = (P4, R2, i, c) }, where:

  R2. (1) double(0) -> 0
      (2) double(s(V)) -> s(s(double(V)))
      (3) minus(X, 0) -> X
      (4) minus(s(Y), s(U)) -> minus(Y, U)

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(Y2, U2) |>| U2

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

Processor output: { }.

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

We use the following projection function:

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

We thus have:

  (1) cons(P2, X3) |>|  X3
  (2) V3           |>=| V3
  (3) X4           |>=| X4

We may remove the strictly oriented DPs.

Processor output: { D8 = (P7, R, i, c) }, where:

  P7. (1) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (2) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** We apply the Usable rules with respect to an argument wiltering [with HORPO] Processor on D7 = (P4, R2, i, c).

Constrained HORPO yields:

  plus#2(s1(P), X1) (>) plus#2(P, X1)
  plus#2(s1(Y1), U1) (>) plus#2(Y1, s1(U1))
  plus#2(s1(V1), W1) (>) plus#2(minus2(V1, W1), double1(W1))
  minus2(X, 00) (>=) X
  minus2(s1(Y), s1(U)) (>=) minus2(Y, U)

We do this using the following settings:

* Disregarded arguments:

  double1 1 
  minus2  2 
  plus#2  2 

* Precedence and permutation:

    double1  { }
  > minus2   { }   1
  = plus#2   { 1 }
  = s1       { 1 }
  > 00       { }

* Well-founded theory orderings:

  [>]_{Bool} = {(true,false)}
  [>]_{Int}  = {(x,y) | x < 1000 /\ x < y }

Processor output: { }.

***** We apply the Graph Processor on D8 = (P7, 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: { }.