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

  Signature: 0       :: b
             1       :: b
             c       :: b -> b
             cons    :: c -> d -> d
             f       :: b -> a
             false   :: a
             filter  :: (c -> a) -> d -> d
             filter2 :: a -> (c -> a) -> c -> d -> d
             g       :: b -> b -> b
             if      :: a -> b -> b -> b
             map     :: (c -> c) -> d -> d
             nil     :: d
             s       :: b -> b
             true    :: a

  Rules: f(0) -> true
         f(1) -> false
         f(s(X)) -> f(X)
         if(true, Y, U) -> Y
         if(false, V, W) -> W
         g(s(P), s(X1)) -> if(f(P), s(P), s(X1))
         g(Y1, c(U1)) -> g(Y1, g(s(c(U1)), U1))
         map(H1, nil) -> nil
         map(I1, cons(P1, X2)) -> cons(I1(P1), map(I1, X2))
         filter(Z2, nil) -> nil
         filter(G2, cons(V2, W2)) -> filter2(G2(V2), G2, V2, W2)
         filter2(true, J2, X3, Y3) -> cons(X3, filter(J2, Y3))
         filter2(false, G3, V3, W3) -> filter(G3, W3)

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) f#(s(X)) => f#(X)
      (2) g#(s(P), s(X1)) => f#(P)
      (3) g#(s(P), s(X1)) => if#(f(P), s(P), s(X1))
      (4) g#(Y1, c(U1)) => g#(s(c(U1)), U1)
      (5) g#(Y1, c(U1)) => g#(Y1, g(s(c(U1)), U1))
      (6) map#(I1, cons(P1, X2)) => map#(I1, X2)
      (7) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (8) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (9) filter2#(false, G3, V3, W3) => filter#(G3, W3)

***** 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: 2 3 4 5 
  5: 2 3 4 5 
  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) f#(s(X)) => f#(X)

  P3. (1) g#(Y1, c(U1)) => g#(s(c(U1)), U1)
      (2) g#(Y1, c(U1)) => g#(Y1, g(s(c(U1)), U1))

  P4. (1) map#(I1, cons(P1, X2)) => map#(I1, X2)

  P5. (1) filter#(G2, cons(V2, W2)) => filter2#(G2(V2), G2, V2, W2)
      (2) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (3) filter2#(false, G3, V3, W3) => filter#(G3, W3)

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

We use the following projection function:

  nu(f#) = 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 Usable Rules Processor on D3 = (P3, R, i, c).

We obtain 7 usable rules (out of 13 rules in the input problem).

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

  R2. (1) f(0) -> true
      (2) f(1) -> false
      (3) f(s(X)) -> f(X)
      (4) g(s(P), s(X1)) -> if(f(P), s(P), s(X1))
      (5) g(Y1, c(U1)) -> g(Y1, g(s(c(U1)), U1))
      (6) if(true, Y, U) -> Y
      (7) if(false, V, W) -> W

***** 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(P1, X2) |>| X2

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(V2, W2) |>|  W2
  (2) Y3           |>=| Y3
  (3) W3           |>=| W3

We may remove the strictly oriented DPs.

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

  P6. (1) filter2#(true, J2, X3, Y3) => filter#(J2, Y3)
      (2) filter2#(false, G3, V3, W3) => filter#(G3, W3)

***** We apply the Usable rules with respect to HORPO Processor on D6 = (P3, R2, i, c).

Constrained HORPO yields:

  g#2(Y1, c1(U1)) (>) g#2(s1(c1(U1)), U1)
  g#2(Y1, c1(U1)) (>) g#2(Y1, g2(s1(c1(U1)), U1))
  g2(s1(P), s1(X1)) (>=) if3(f1(P), s1(P), s1(X1))
  g2(Y1, c1(U1)) (>=) g2(Y1, g2(s1(c1(U1)), U1))
  if3(true0, Y, U) (>=) Y
  if3(false0, V, W) (>=) W

We do this using the following settings:

* Disregarded arguments:

  g#2 1 
  if3 1 
  s1  1 

* Precedence and permutation:

    00      { }
  = 10      { }
  = c1      { 1 }
  = s1      { }
  > f1      { 1 }
  = false0  { }
  = g#2     { 2 }
  = g2      { 2 } 1
  = if3     { 3 } 2
  = true0   { }

* Well-founded theory orderings:

  [>]_{Bool} = {(true,false)}
  [>]_{Int}  = {(x,y) | x < 1000 /\ x < y }

Processor output: { }.

***** We apply the Graph Processor on D7 = (P6, R, i, c).

We compute a graph approximation with the following edges:

  1:
  2:

As there are no SCCs, this DP problem is removed.

Processor output: { }.