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

  Signature: 0     :: c
             cons  :: a -> b -> b
             div   :: c -> c -> c
             map   :: (a -> a) -> b -> b
             minus :: c -> c -> c
             nil   :: b
             s     :: c -> c

  Rules: map(F, nil) -> nil
         map(Z, cons(U, V)) -> cons(Z(U), map(Z, V))
         minus(W, 0) -> W
         minus(s(P), s(X1)) -> minus(P, X1)
         div(0, s(Y1)) -> 0
         div(s(U1), s(V1)) -> s(div(minus(U1, V1), s(V1)))

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) minus#(s(P), s(X1)) => minus#(P, X1)
      (3) div#(s(U1), s(V1)) => minus#(U1, V1)
      (4) div#(s(U1), s(V1)) => div#(minus(U1, V1), s(V1))

***** 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 

There are 3 SCCs.

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

  P2. (1) map#(Z, cons(U, V)) => map#(Z, V)

  P3. (1) minus#(s(P), s(X1)) => minus#(P, X1)

  P4. (1) div#(s(U1), s(V1)) => div#(minus(U1, V1), s(V1))

***** 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(minus#) = 1

We thus have:

  (1) s(P) |>| P

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

Processor output: { }.

***** We apply the Reduction Pair [with HORPO] Processor on D4 = (P4, R, i, c).

Constrained HORPO yields:

  div#(s(U1), s(V1)) (>) div#(minus(U1, V1), s(V1))
  map(F, nil) (>=) nil
  map(Z, cons(U, V)) (>=) cons(Z(U), map(Z, V))
  minus(W, 0) (>=) W
  minus(s(P), s(X1)) (>=) minus(P, X1)
  div(0, s(Y1)) (>=) 0
  div(s(U1), s(V1)) (>=) s(div(minus(U1, V1), s(V1)))

We do this using the following settings:

* Disregarded arguments:

  cons  1 
  div#  2 
  div   2 
  minus 2 

* Precedence and permutation:

    div    { }   1
  = nil    { }
  > s      { 1 }
  > div#   { 1 }
  = map    { 1 } 2
  > minus  { 1 }
  > 0      { }
  = cons   { 2 }

* Well-founded theory orderings:

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

All dependency pairs were removed.

Processor output: { }.