Term Rewriting System R:
[X]
f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(0)) -> F(p(s(0)))
F(s(0)) -> P(s(0))
ACTIVATE(nf(X)) -> F(activate(X))
ACTIVATE(nf(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(n0) -> 0'

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS


Dependency Pair:

F(s(0)) -> F(p(s(0)))


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





The following dependency pair can be strictly oriented:

F(s(0)) -> F(p(s(0)))


The following rules can be oriented:

p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{activate, 0, p} > {ns, f, s} > nf
{activate, 0, p} > {ns, f, s} > cons
{activate, 0, p} > n0

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
s(x1) -> s(x1)
p(x1) -> p
0 -> 0
ns(x1) -> ns(x1)
f(x1) -> f(x1)
cons(x1, x2) -> cons(x1, x2)
nf(x1) -> nf(x1)
activate(x1) -> activate(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 3
Dependency Graph
       →DP Problem 2
AFS


Dependency Pair:


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Argument Filtering and Ordering


Dependency Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nf(X)) -> ACTIVATE(X)


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





The following dependency pairs can be strictly oriented:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nf(X)) -> ACTIVATE(X)


The following rules can be oriented:

f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{activate, 0} > f > nf
{activate, 0} > f > cons
{activate, 0} > n0
{activate, 0} > s > ns

resulting in one new DP problem.
Used Argument Filtering System:
ACTIVATE(x1) -> ACTIVATE(x1)
ns(x1) -> ns(x1)
nf(x1) -> nf(x1)
f(x1) -> f(x1)
cons(x1, x2) -> cons(x1, x2)
0 -> 0
s(x1) -> s(x1)
p(x1) -> x1
activate(x1) -> activate(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 4
Dependency Graph


Dependency Pair:


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes