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

  Signature: append  :: list -> list -> list
             cons    :: nat -> list -> list
             map     :: (nat -> nat) -> list -> list
             mirror  :: list -> list
             nil     :: list
             reverse :: list -> list
             shuffle :: list -> list

  Rules: 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))

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) 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 D1 = (P1, R, i, c).

We compute a graph approximation with the following edges:

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

There are 4 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) ; D5 = (P5, R, i, c) }, where:

  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 D2 = (P2, R, i, c).

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.

Processor output: { }.

***** We apply the Reduction Pair [with HORPO] Processor on D3 = (P3, R, i, c).

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:

* Disregarded arguments:

  cons 1 

* Precedence and permutation:

    map       { 2 } 1
  > mirror    { 1 }
  = shuffle   { 1 }
  > append    { 2 } 1
  = cons      { }   2
  = nil       { }
  > reverse   { 1 }
  = shuffle#  { 1 }

* Well-founded theory orderings:

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

All dependency pairs were removed.

Processor output: { }.

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

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.

Processor output: { }.

***** 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(X, Y) |>| Y

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

Processor output: { }.