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

  Signature: cons      :: a -> c -> c
             dropWhile :: (a -> b) -> c -> c
             if        :: b -> c -> c -> c
             nil       :: c
             takeWhile :: (a -> b) -> c -> c
             true      :: b

  Rules: if(true, X, Y) -> X
         if(true, U, V) -> V
         takeWhile(I, nil) -> nil
         takeWhile(J, cons(X1, Y1)) -> if(J(X1), cons(X1, takeWhile(J, Y1)), nil)
         dropWhile(G1, nil) -> nil
         dropWhile(H1, cons(W1, P1)) -> if(H1(W1), dropWhile(H1, P1), cons(W1, 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, i, c), where:

  P1. (1) takeWhile#(J, cons(X1, Y1)) => takeWhile#(J, Y1)
      (2) takeWhile#(J, cons(X1, Y1)) => if#(J(X1), cons(X1, takeWhile(J, Y1)), nil)
      (3) dropWhile#(H1, cons(W1, P1)) => dropWhile#(H1, P1)
      (4) dropWhile#(H1, cons(W1, P1)) => if#(H1(W1), dropWhile(H1, P1), cons(W1, P1))

***** 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 4 
  4:

There are 2 SCCs.

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

  P2. (1) takeWhile#(J, cons(X1, Y1)) => takeWhile#(J, Y1)

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

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

We use the following projection function:

  nu(takeWhile#) = 2

We thus have:

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

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(dropWhile#) = 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: { }.