Term Rewriting System R:
[x, y]
f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(s(x))) -> P(h(g(x)))
F(s(s(x))) -> H(g(x))
F(s(s(x))) -> G(x)
F(s(s(x))) -> +'(p(g(x)), q(g(x)))
F(s(s(x))) -> P(g(x))
F(s(s(x))) -> Q(g(x))
G(s(x)) -> H(g(x))
G(s(x)) -> G(x)
G(s(x)) -> +'(p(g(x)), q(g(x)))
G(s(x)) -> P(g(x))
G(s(x)) -> Q(g(x))
H(x) -> +'(p(x), q(x))
H(x) -> P(x)
H(x) -> Q(x)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pair:

+'(x, s(y)) -> +'(x, y)


Rules:


f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))





The following dependency pair can be strictly oriented:

+'(x, s(y)) -> +'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


Rules:


f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pair:

G(s(x)) -> G(x)


Rules:


f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))





The following dependency pair can be strictly oriented:

G(s(x)) -> G(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(G(x1))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph


Dependency Pair:


Rules:


f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))





Using the Dependency Graph resulted in no new DP problems.

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