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

  Signature: 0       :: b
             cons    :: c -> d -> d
             false   :: a
             filter  :: (c -> a) -> d -> d
             filter2 :: a -> (c -> a) -> c -> d -> d
             if      :: a -> b -> b -> b
             le      :: b -> b -> a
             map     :: (c -> c) -> d -> d
             minus   :: b -> b -> b
             nil     :: d
             p       :: b -> b
             s       :: b -> b
             true    :: a

  Rules: p(0) -> 0
         p(s(X)) -> X
         le(0, Y) -> true
         le(s(U), 0) -> false
         le(s(V), s(W)) -> le(V, W)
         minus(P, X1) -> if(le(P, X1), P, X1)
         if(true, Y1, U1) -> 0
         if(false, V1, W1) -> s(minus(p(V1), W1))
         map(J1, nil) -> nil
         map(F2, cons(Y2, U2)) -> cons(F2(Y2), map(F2, U2))
         filter(H2, nil) -> nil
         filter(I2, cons(P2, X3)) -> filter2(I2(P2), I2, P2, X3)
         filter2(true, Z3, U3, V3) -> cons(U3, filter(Z3, V3))
         filter2(false, I3, P3, X4) -> filter(I3, X4)

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) le#(s(V), s(W)) => le#(V, W)
      (2) minus#(P, X1) => le#(P, X1)
      (3) minus#(P, X1) => if#(le(P, X1), P, X1)
      (4) if#(false, V1, W1) => p#(V1)
      (5) if#(false, V1, W1) => minus#(p(V1), W1)
      (6) map#(F2, cons(Y2, U2)) => map#(F2, U2)
      (7) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (8) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (9) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

We compute a graph approximation with the following edges:

  1: 1 
  2: 1 
  3: 4 5 
  4:
  5: 2 3 
  6: 6 
  7: 8 9 
  8: 7 
  9: 7 

There are 4 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) ; D5 = (P5, R, i, c) }, where:

  P2. (1) le#(s(V), s(W)) => le#(V, W)

  P3. (1) minus#(P, X1) => if#(le(P, X1), P, X1)
      (2) if#(false, V1, W1) => minus#(p(V1), W1)

  P4. (1) map#(F2, cons(Y2, U2)) => map#(F2, U2)

  P5. (1) filter#(I2, cons(P2, X3)) => filter2#(I2(P2), I2, P2, X3)
      (2) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (3) filter2#(false, I3, P3, X4) => filter#(I3, X4)

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

We use the following projection function:

  nu(le#) = 1

We thus have:

  (1) s(V) |>| V

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

Processor output: { }.

***** We apply the Usable Rules Processor on D3 = (P3, R, i, c).

We obtain 5 usable rules (out of 14 rules in the input problem).

Processor output: { D6 = (P3, R2, i, c) }, where:

  R2. (1) le(0, Y) -> true
      (2) le(s(U), 0) -> false
      (3) le(s(V), s(W)) -> le(V, W)
      (4) p(0) -> 0
      (5) p(s(X)) -> X

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

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(Y2, U2) |>| U2

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

Processor output: { }.

***** We apply the Subterm Criterion Processor on D5 = (P5, R, i, c).

We use the following projection function:

  nu(filter#)  = 2
  nu(filter2#) = 4

We thus have:

  (1) cons(P2, X3) |>|  X3
  (2) V3           |>=| V3
  (3) X4           |>=| X4

We may remove the strictly oriented DPs.

Processor output: { D7 = (P6, R, i, c) }, where:

  P6. (1) filter2#(true, Z3, U3, V3) => filter#(Z3, V3)
      (2) filter2#(false, I3, P3, X4) => filter#(I3, X4)

***** No progress could be made on DP problem D6 = (P3, R2, i, c).