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

  Signature: cons  :: nat -> list -> list
             map   :: (nat -> nat) -> list -> list
             merge :: list -> list -> list -> list
             nil   :: list

  Rules: merge(nil, nil, X) -> X
         merge(nil, cons(Y, U), V) -> merge(U, nil, cons(Y, V))
         merge(cons(W, P), X1, Y1) -> merge(X1, P, cons(W, Y1))
         map(G1, nil) -> nil
         map(H1, cons(W1, P1)) -> cons(H1(W1), map(H1, P1))

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, f, c), where:

  P1. (1) merge#(nil, cons(Y, U), V) => merge#(U, nil, cons(Y, V))
      (2) merge#(cons(W, P), X1, Y1) => merge#(X1, P, cons(W, Y1))
      (3) map#(H1, cons(W1, P1)) => map#(H1, P1)

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

We compute a graph approximation with the following edges:

  1: 2 
  2: 1 2 
  3: 3 

There are 2 SCCs.

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

  P2. (1) merge#(nil, cons(Y, U), V) => merge#(U, nil, cons(Y, V))
      (2) merge#(cons(W, P), X1, Y1) => merge#(X1, P, cons(W, Y1))

  P3. (1) map#(H1, cons(W1, P1)) => map#(H1, P1)

***** We apply the Reduction Pair [with HORPO] Processor on D2 = (P2, R, f, c).

Constrained HORPO yields:

  merge#(nil, cons(Y, U), V) (>) merge#(U, nil, cons(Y, V))
  merge#(cons(W, P), X1, Y1) (>) merge#(X1, P, cons(W, Y1))
  merge(nil, nil, X) (>=) X
  merge(nil, cons(Y, U), V) (>=) merge(U, nil, cons(Y, V))
  merge(cons(W, P), X1, Y1) (>=) merge(X1, P, cons(W, Y1))
  map(G1, nil) (>=) nil
  map(H1, cons(W1, P1)) (>=) cons(H1(W1), map(H1, P1))

We do this using the following settings:

* Disregarded arguments:

  cons   1 
  map    1 
  merge# 3 

* Precedence and permutation:

    cons    { }     2
  = map     { }     2
  = merge   { 1 2 } 3
  = nil     { }
  > merge#  { 1 2 }

* 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 D3 = (P3, R, f, c).

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(W1, P1) |>| P1

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

Processor output: { }.