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

  Signature: 0      :: a
             eq     :: a -> a -> c
             false  :: c
             fork   :: b -> a -> b -> b
             if     :: c -> c -> c -> c
             lt     :: a -> a -> c
             member :: a -> b -> c
             null   :: b
             s      :: a -> a
             true   :: c

  Rules: lt(s(X), s(Y)) -> lt(X, Y)
         lt(0, s(U)) -> true
         lt(V, 0) -> false
         eq(W, W) -> true
         eq(s(P), 0) -> false
         eq(0, s(X1)) -> false
         member(Y1, null) -> false
         member(U1, fork(V1, W1, P1)) -> if(lt(U1, W1), member(U1, V1), if(eq(U1, W1), true, member(U1, 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, f, c), where:

  P1. (1) lt#(s(X), s(Y)) => lt#(X, Y)
      (2) member#(U1, fork(V1, W1, P1)) => lt#(U1, W1)
      (3) member#(U1, fork(V1, W1, P1)) => member#(U1, V1)
      (4) member#(U1, fork(V1, W1, P1)) => eq#(U1, W1)
      (5) member#(U1, fork(V1, W1, P1)) => member#(U1, P1)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 1 
  3: 2 3 4 5 
  4:
  5: 2 3 4 5 

There are 2 SCCs.

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

  P2. (1) lt#(s(X), s(Y)) => lt#(X, Y)

  P3. (1) member#(U1, fork(V1, W1, P1)) => member#(U1, V1)
      (2) member#(U1, fork(V1, W1, P1)) => member#(U1, P1)

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

We use the following projection function:

  nu(lt#) = 1

We thus have:

  (1) s(X) |>| X

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(member#) = 2

We thus have:

  (1) fork(V1, W1, P1) |>| V1
  (2) fork(V1, W1, P1) |>| P1

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

Processor output: { }.