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

  Signature: 0       :: b
             cons    :: b -> b -> b
             del     :: b -> b -> b
             eq      :: b -> b -> a
             false   :: a
             filter  :: (b -> a) -> b -> b
             filter2 :: a -> (b -> a) -> b -> b -> b
             if      :: a -> b -> b -> b
             le      :: b -> b -> a
             map     :: (b -> b) -> b -> b
             min     :: b -> b -> b
             minsort :: b -> b
             nil     :: b
             s       :: b -> b
             true    :: a

  Rules: le(0, X) -> true
         le(s(Y), 0) -> false
         le(s(U), s(V)) -> le(U, V)
         eq(0, 0) -> true
         eq(0, s(W)) -> false
         eq(s(P), 0) -> false
         eq(s(X1), s(Y1)) -> eq(X1, Y1)
         if(true, U1, V1) -> U1
         if(false, W1, P1) -> P1
         minsort(nil) -> nil
         minsort(cons(X2, Y2)) -> cons(min(X2, Y2), minsort(del(min(X2, Y2), cons(X2, Y2))))
         min(U2, nil) -> U2
         min(V2, cons(W2, P2)) -> if(le(V2, W2), min(V2, P2), min(W2, P2))
         del(X3, nil) -> nil
         del(Y3, cons(U3, V3)) -> if(eq(Y3, U3), V3, cons(U3, del(Y3, V3)))
         map(I3, nil) -> nil
         map(J3, cons(X4, Y4)) -> cons(J3(X4), map(J3, Y4))
         filter(G4, nil) -> nil
         filter(H4, cons(W4, P4)) -> filter2(H4(W4), H4, W4, P4)
         filter2(true, F5, Y5, U5) -> cons(Y5, filter(F5, U5))
         filter2(false, H5, W5, P5) -> filter(H5, P5)

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(U), s(V)) => le#(U, V)
      (2)  eq#(s(X1), s(Y1)) => eq#(X1, Y1)
      (3)  minsort#(cons(X2, Y2)) => min#(X2, Y2)
      (4)  minsort#(cons(X2, Y2)) => min#(X2, Y2)
      (5)  minsort#(cons(X2, Y2)) => del#(min(X2, Y2), cons(X2, Y2))
      (6)  minsort#(cons(X2, Y2)) => minsort#(del(min(X2, Y2), cons(X2, Y2)))
      (7)  min#(V2, cons(W2, P2)) => le#(V2, W2)
      (8)  min#(V2, cons(W2, P2)) => min#(V2, P2)
      (9)  min#(V2, cons(W2, P2)) => min#(W2, P2)
      (10) min#(V2, cons(W2, P2)) => if#(le(V2, W2), min(V2, P2), min(W2, P2))
      (11) del#(Y3, cons(U3, V3)) => eq#(Y3, U3)
      (12) del#(Y3, cons(U3, V3)) => del#(Y3, V3)
      (13) del#(Y3, cons(U3, V3)) => if#(eq(Y3, U3), V3, cons(U3, del(Y3, V3)))
      (14) map#(J3, cons(X4, Y4)) => map#(J3, Y4)
      (15) filter#(H4, cons(W4, P4)) => filter2#(H4(W4), H4, W4, P4)
      (16) filter2#(true, F5, Y5, U5) => filter#(F5, U5)
      (17) filter2#(false, H5, W5, P5) => filter#(H5, P5)

***** 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:  7 8 9 10 
  4:  7 8 9 10 
  5:  11 12 13 
  6:  3 4 5 6 
  7:  1 
  8:  7 8 9 10 
  9:  7 8 9 10 
  10:
  11: 2 
  12: 11 12 13 
  13:
  14: 14 
  15: 16 17 
  16: 15 
  17: 15 

There are 7 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) ; D5 = (P5, R, i, c) ; D6 = (P6, R, i, c) ; D7 = (P7, R, i, c) ; D8 = (P8, R, i, c) }, where:

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

  P3. (1) eq#(s(X1), s(Y1)) => eq#(X1, Y1)

  P4. (1) min#(V2, cons(W2, P2)) => min#(V2, P2)
      (2) min#(V2, cons(W2, P2)) => min#(W2, P2)

  P5. (1) del#(Y3, cons(U3, V3)) => del#(Y3, V3)

  P6. (1) minsort#(cons(X2, Y2)) => minsort#(del(min(X2, Y2), cons(X2, Y2)))

  P7. (1) map#(J3, cons(X4, Y4)) => map#(J3, Y4)

  P8. (1) filter#(H4, cons(W4, P4)) => filter2#(H4(W4), H4, W4, P4)
      (2) filter2#(true, F5, Y5, U5) => filter#(F5, U5)
      (3) filter2#(false, H5, W5, P5) => filter#(H5, P5)

***** 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(U) |>| U

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(eq#) = 1

We thus have:

  (1) s(X1) |>| X1

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

Processor output: { }.

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

We use the following projection function:

  nu(min#) = 2

We thus have:

  (1) cons(W2, P2) |>| P2
  (2) cons(W2, P2) |>| P2

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(del#) = 2

We thus have:

  (1) cons(U3, V3) |>| V3

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

Processor output: { }.

***** We apply the Usable Rules Processor on D6 = (P6, R, i, c).

We obtain 13 usable rules (out of 21 rules in the input problem).

Processor output: { D9 = (P6, R2, i, c) }, where:

  R2. (1)  del(X3, nil) -> nil
      (2)  del(Y3, cons(U3, V3)) -> if(eq(Y3, U3), V3, cons(U3, del(Y3, V3)))
      (3)  eq(0, 0) -> true
      (4)  eq(0, s(W)) -> false
      (5)  eq(s(P), 0) -> false
      (6)  eq(s(X1), s(Y1)) -> eq(X1, Y1)
      (7)  if(true, U1, V1) -> U1
      (8)  if(false, W1, P1) -> P1
      (9)  le(0, X) -> true
      (10) le(s(Y), 0) -> false
      (11) le(s(U), s(V)) -> le(U, V)
      (12) min(U2, nil) -> U2
      (13) min(V2, cons(W2, P2)) -> if(le(V2, W2), min(V2, P2), min(W2, P2))

***** We apply the Subterm Criterion Processor on D7 = (P7, R, i, c).

We use the following projection function:

  nu(map#) = 2

We thus have:

  (1) cons(X4, Y4) |>| Y4

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

Processor output: { }.

***** We apply the Subterm Criterion Processor on D8 = (P8, R, i, c).

We use the following projection function:

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

We thus have:

  (1) cons(W4, P4) |>|  P4
  (2) U5           |>=| U5
  (3) P5           |>=| P5

We may remove the strictly oriented DPs.

Processor output: { D10 = (P9, R, i, c) }, where:

  P9. (1) filter2#(true, F5, Y5, U5) => filter#(F5, U5)
      (2) filter2#(false, H5, W5, P5) => filter#(H5, P5)

***** No progress could be made on DP problem D9 = (P6, R2, i, c).