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

  Signature: cons  :: d -> e -> e
             g     :: b
             map_1 :: (d -> d) -> e -> e
             map_2 :: (d -> a -> d) -> a -> e -> e
             map_3 :: (b -> d -> c -> d) -> b -> c -> e -> e

  Rules: map_1(F, cons(Y, U)) -> cons(F(Y), map_1(F, U))
         map_2(I, V, cons(P, X1)) -> cons(I(P, V), map_2(I, V, X1))
         map_3(G1, g, Y1, cons(V1, W1)) -> cons(G1(g, V1, Y1), map_3(G1, g, Y1, W1))

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) map_1#(F, cons(Y, U)) => map_1#(F, U)
      (2) map_2#(I, V, cons(P, X1)) => map_2#(I, V, X1)
      (3) map_3#(G1, g, Y1, cons(V1, W1)) => map_3#(G1, g, Y1, W1)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 2 
  3: 3 

There are 3 SCCs.

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

  P2. (1) map_1#(F, cons(Y, U)) => map_1#(F, U)

  P3. (1) map_2#(I, V, cons(P, X1)) => map_2#(I, V, X1)

  P4. (1) map_3#(G1, g, Y1, cons(V1, W1)) => map_3#(G1, g, Y1, W1)

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

We use the following projection function:

  nu(map_1#) = 2

We thus have:

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

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(map_2#) = 3

We thus have:

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

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

Processor output: { }.

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

We use the following projection function:

  nu(map_3#) = 4

We thus have:

  (1) cons(V1, W1) |>| W1

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

Processor output: { }.