Term Rewriting System R:
[X, Y, Z, X1, X2]
active(f(X)) -> mark(cons(X, f(g(X))))
active(g(0)) -> mark(s(0))
active(g(s(X))) -> mark(s(s(g(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(g(X)) -> g(active(X))
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(f(X)) -> f(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(g(X)) -> g(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ACTIVE(f(X)) -> CONS(X, f(g(X)))
ACTIVE(f(X)) -> F(g(X))
ACTIVE(f(X)) -> G(X)
ACTIVE(g(0)) -> S(0)
ACTIVE(g(s(X))) -> S(s(g(X)))
ACTIVE(g(s(X))) -> S(g(X))
ACTIVE(g(s(X))) -> G(X)
ACTIVE(sel(s(X), cons(Y, Z))) -> SEL(X, Z)
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(g(X)) -> G(active(X))
ACTIVE(g(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(sel(X1, X2)) -> SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) -> ACTIVE(X1)
ACTIVE(sel(X1, X2)) -> SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) -> ACTIVE(X2)
F(mark(X)) -> F(X)
F(ok(X)) -> F(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
G(mark(X)) -> G(X)
G(ok(X)) -> G(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
SEL(mark(X1), X2) -> SEL(X1, X2)
SEL(X1, mark(X2)) -> SEL(X1, X2)
SEL(ok(X1), ok(X2)) -> SEL(X1, X2)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(g(X)) -> G(proper(X))
PROPER(g(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(sel(X1, X2)) -> SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) -> PROPER(X1)
PROPER(sel(X1, X2)) -> PROPER(X2)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains eight SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)


Rules:


active(f(X)) -> mark(cons(X, f(g(X))))
active(g(0)) -> mark(s(0))
active(g(s(X))) -> mark(s(s(g(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(g(X)) -> g(active(X))
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(f(X)) -> f(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(g(X)) -> g(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)


Additionally, the following rules can be oriented:

active(f(X)) -> mark(cons(X, f(g(X))))
active(g(0)) -> mark(s(0))
active(g(s(X))) -> mark(s(s(g(X))))
active(sel(0, cons(X, Y))) -> mark(X)
active(sel(s(X), cons(Y, Z))) -> mark(sel(X, Z))
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(g(X)) -> g(active(X))
active(s(X)) -> s(active(X))
active(sel(X1, X2)) -> sel(active(X1), X2)
active(sel(X1, X2)) -> sel(X1, active(X2))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
sel(mark(X1), X2) -> mark(sel(X1, X2))
sel(X1, mark(X2)) -> mark(sel(X1, X2))
sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
proper(f(X)) -> f(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(g(X)) -> g(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(proper(x1))=  1  
  POL(0)=  0  
  POL(g(x1))=  x1  
  POL(cons(x1, x2))=  x2  
  POL(s(x1))=  x1  
  POL(sel(x1, x2))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(f(x1))=  x1  
  POL(CONS(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)
       →DP Problem 6
         ↳Remaining Obligation(s)
       →DP Problem 7
         ↳Remaining Obligation(s)
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:

Termination of R could not be shown.
Duration:
0:00 minutes