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)  reverse#(V) => reverse2#(V, nil)
      (2)  reverse2#(cons(X1, Y1), P) => reverse2#(Y1, cons(X1, P))
      (3)  last#(X{12}) => hd#(X{13})
      (4)  last#(X{12}) => reverse#(X{14})
      (5)  last#(X{12}) => compose#(hd, reverse, X{12})
      (6)  init#(X{15}) => reverse#(X{16})
      (7)  init#(X{15}) => tl#(X{17})
      (8)  init#(X{15}) => reverse#(X{18})
      (9)  init#(X{15}) => compose#(tl, reverse, X{19})
      (10) init#(X{15}) => compose#(reverse, compose(tl, reverse), X{15})

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

There is only one SCC, so all DPs not inside the SCC can be removed:

  P2. (1) reverse2#(cons(X1, Y1), P) => reverse2#(Y1, cons(X1, P))

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

We use the following projection function:

  nu(reverse2#) = 1

We thus have:

  (1) cons(X1, Y1) |>| Y1

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