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

  Signature: and     :: c -> c -> c
             cons    :: a -> b -> b
             false   :: c
             forall  :: (a -> c) -> b -> c
             forsome :: (a -> c) -> b -> c
             nil     :: b
             or      :: c -> c -> c
             true    :: c

  Rules: and(true, true) -> true
         and(true, false) -> false
         and(false, true) -> false
         and(false, false) -> false
         or(true, true) -> true
         or(true, false) -> true
         or(false, true) -> true
         or(false, false) -> false
         forall(F, nil) -> true
         forall(Z, cons(U, V)) -> and(Z(U), forall(Z, V))
         forsome(I, nil) -> false
         forsome(J, cons(X1, Y1)) -> or(J(X1), forsome(J, Y1))

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) forall#(Z, cons(U, V)) => forall#(Z, V)
      (2) forall#(Z, cons(U, V)) => and#(Z(U), forall(Z, V))
      (3) forsome#(J, cons(X1, Y1)) => forsome#(J, Y1)
      (4) forsome#(J, cons(X1, Y1)) => or#(J(X1), forsome(J, Y1))

***** 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) forall#(Z, cons(U, V)) => forall#(Z, V)

  P3. (1) forsome#(J, cons(X1, Y1)) => forsome#(J, Y1)

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

We use the following projection function:

  nu(forall#) = 2

We thus have:

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

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(forsome#) = 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: { }.