Term Rewriting System R:
[x]
f(s(x)) -> s(f(f(p(s(x)))))
f(0) -> 0
p(s(x)) -> x

Termination of R to be shown. 



   TRS
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(x)) -> F(f(p(s(x))))
F(s(x)) -> F(p(s(x)))
F(s(x)) -> P(s(x))

Furthermore, R contains one SCC. 


   TRS
     ↳DPs
       →DP Problem 1
         ↳Non-Overlappingness Check


Dependency Pairs:

F(s(x)) -> F(p(s(x)))
F(s(x)) -> F(f(p(s(x))))


Rules:


f(s(x)) -> s(f(f(p(s(x)))))
f(0) -> 0
p(s(x)) -> x





R does not overlap into P. Moreover, R is locally confluent (all critical pairs are trivially joinable).Hence we can switch to innermost.
The transformation is resulting in one subcycle: 


   TRS
     ↳DPs
       →DP Problem 1
         ↳NOC
           →DP Problem 2
             ↳Rewriting Transformation


Dependency Pairs:

F(s(x)) -> F(p(s(x)))
F(s(x)) -> F(f(p(s(x))))


Rules:


f(s(x)) -> s(f(f(p(s(x)))))
f(0) -> 0
p(s(x)) -> x


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

F(s(x)) -> F(f(p(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(f(x))

The transformation is resulting in one new DP problem: 



   TRS
     ↳DPs
       →DP Problem 1
         ↳NOC
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 3
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(x)) -> F(f(x))
F(s(x)) -> F(p(s(x)))


Rules:


f(s(x)) -> s(f(f(p(s(x)))))
f(0) -> 0
p(s(x)) -> x


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

F(s(x)) -> F(p(s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(x)

The transformation is resulting in one new DP problem: 



   TRS
     ↳DPs
       →DP Problem 1
         ↳NOC
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 4
                 ↳Negative Polynomial Order


Dependency Pairs:

F(s(x)) -> F(x)
F(s(x)) -> F(f(x))


Rules:


f(s(x)) -> s(f(f(p(s(x)))))
f(0) -> 0
p(s(x)) -> x


Strategy:

innermost




The following Dependency Pairs can be strictly oriented using the given order.

F(s(x)) -> F(x)
F(s(x)) -> F(f(x))


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

f(0) -> 0
p(s(x)) -> x
f(s(x)) -> s(f(f(p(s(x)))))


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1) ) = x1

POL( s(x1) ) = x1 + 1

POL( f(x1) ) = x1

POL( 0 ) = 0

POL( p(x1) ) = max{0, x1 - 1}


This results in one new DP problem. 


   TRS
     ↳DPs
       →DP Problem 1
         ↳NOC
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(s(x)) -> s(f(f(p(s(x)))))
f(0) -> 0
p(s(x)) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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