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

  Signature: 0     :: a
             comp  :: (b -> b) -> (b -> b) -> b -> b
             plus  :: a -> a -> a
             s     :: a -> a
             times :: a -> a -> a
             twice :: (b -> b) -> b -> b

  Rules: plus(0, X) -> X
         plus(s(Y), U) -> s(plus(Y, U))
         times(0, V) -> 0
         times(s(W), P) -> plus(times(W, P), P)
         comp(F1, Z1, U1) -> F1(Z1(U1))
         twice(H1) -> comp(H1, H1)

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) plus#(s(Y), U) => plus#(Y, U)
      (2) times#(s(W), P) => times#(W, P)
      (3) times#(s(W), P) => plus#(times(W, P), P)
      (4) twice#(H1, arg2) => comp#(H1, H1, arg2)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 2 3 
  3: 1 
  4:

There are 2 SCCs.

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

  P2. (1) plus#(s(Y), U) => plus#(Y, U)

  P3. (1) times#(s(W), P) => times#(W, P)

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

We use the following projection function:

  nu(plus#) = 1

We thus have:

  (1) s(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 D3 = (P3, R, f, c).

We use the following projection function:

  nu(times#) = 1

We thus have:

  (1) s(W) |>| W

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

Processor output: { }.