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

  Signature: cons    :: nat -> list -> list
             explode :: list -> (nat -> nat) -> nat -> nat
             implode :: list -> (nat -> nat) -> nat -> nat
             nil     :: list
             op      :: (nat -> nat) -> (nat -> nat) -> nat -> nat

  Rules: op(F, G, X) -> F(G(X))
         implode(nil, F, X) -> X
         implode(cons(H, T), F, X) -> implode(T, F, F(X))
         explode(nil, F, X) -> X
         explode(cons(H, T), F, X) -> explode(T, op(F, F), F(X))

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) implode#(cons(H, T), F, X) => implode#(T, F, F(X))
      (2) explode#(cons(H, T), F, X) => op#(F, F, fresh1)
      (3) explode#(cons(H, T), F, X) => explode#(T, op(F, F), F(X))

***** 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 

There are 2 SCCs.

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

  P2. (1) implode#(cons(H, T), F, X) => implode#(T, F, F(X))

  P3. (1) explode#(cons(H, T), F, X) => explode#(T, op(F, F), F(X))

***** We apply the Subterm Criterion Processor on D2 = (P2, R, i, c).

We use the following projection function:

  nu(implode#) = 1

We thus have:

  (1) cons(H, T) |>| T

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(explode#) = 1

We thus have:

  (1) cons(H, T) |>| T

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

Processor output: { }.