The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the following initial DP problem:

  P1. (1) append#(cons(X, Y), U) => append#(Y, U)
      (2) shuffle#(cons(X, Y)) => reverse#(Y)
      (3) shuffle#(cons(X, Y)) => shuffle#(reverse(Y))
      (4) mirror#(cons(X, Y)) => mirror#(Y)
      (5) mirror#(cons(X, Y)) => append#(cons(X, mirror(Y)), cons(X, nil))
      (6) map#(H, cons(X, Y)) => map#(H, Y)

***** We apply the Graph Processor on P1.

Considering the 4 SCCs, this DP problem is split into the following new problems.

  P2. (1) append#(cons(X, Y), U) => append#(Y, U)

  P3. (1) shuffle#(cons(X, Y)) => shuffle#(reverse(Y))

  P4. (1) mirror#(cons(X, Y)) => mirror#(Y)

  P5. (1) map#(H, cons(X, Y)) => map#(H, Y)

***** We apply the Subterm Criterion Processor on P2.

We use the following projection function:

  nu(append#) = 1

We thus have:

  (1) cons(X, Y) |>| Y

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

***** We apply the horpo Processor on P3.

Constrained HORPO yields:

  shuffle#(cons(X, Y)) (>) shuffle#(reverse(Y))
  append(nil, U) (>=) U
  append(cons(X, Y), U) (>=) cons(X, append(Y, U))
  reverse(nil) (>=) nil
  shuffle(nil) (>=) nil
  shuffle(cons(X, Y)) (>=) cons(X, shuffle(reverse(Y)))
  mirror(nil) (>=) nil
  mirror(cons(X, Y)) (>=) append(cons(X, mirror(Y)), cons(X, nil))
  map(H, nil) (>=) nil
  map(H, cons(X, Y)) (>=) cons(H(X), map(H, Y))

We do this using the following settings:

* Precedence and status (for non-mentioned symbols the precedence is irrelevant and the status is Lex):

  cons               (status: Lex)   >
  reverse = shuffle  (status: Lex)   >
  map                (status: Lex)   >
  nil                (status: Lex)   >
  shuffle#           (status: Lex)   >
  mirror             (status: Lex)   >
  append             (status: Mul_2)

* Well-founded theory orderings:

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

* Filter:

  cons    disregards argument(s) 1 2 
  map     disregards argument(s) 1 
  reverse disregards argument(s) 1 

All dependency pairs were removed.

***** We apply the Subterm Criterion Processor on P4.

We use the following projection function:

  nu(mirror#) = 1

We thus have:

  (1) cons(X, Y) |>| Y

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

***** We apply the Subterm Criterion Processor on P5.

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(X, Y) |>| Y

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