Term Rewriting System R:
[x]
sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

SUM(s(x)) -> SUM(x)
SUM1(s(x)) -> SUM1(x)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

SUM(s(x)) -> SUM(x)


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




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

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

SUM(s(s(x''))) -> SUM(s(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

SUM(s(s(x''))) -> SUM(s(x''))


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




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

SUM(s(s(x''))) -> SUM(s(x''))
one new Dependency Pair is created:

SUM(s(s(s(x'''')))) -> SUM(s(s(x'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

SUM(s(s(s(x'''')))) -> SUM(s(s(x'''')))


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

SUM(s(s(s(x'''')))) -> SUM(s(s(x'''')))


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
SUM(x1) -> SUM(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Forward Instantiation Transformation


Dependency Pair:

SUM1(s(x)) -> SUM1(x)


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




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

SUM1(s(x)) -> SUM1(x)
one new Dependency Pair is created:

SUM1(s(s(x''))) -> SUM1(s(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳Forward Instantiation Transformation


Dependency Pair:

SUM1(s(s(x''))) -> SUM1(s(x''))


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




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

SUM1(s(s(x''))) -> SUM1(s(x''))
one new Dependency Pair is created:

SUM1(s(s(s(x'''')))) -> SUM1(s(s(x'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Argument Filtering and Ordering


Dependency Pair:

SUM1(s(s(s(x'''')))) -> SUM1(s(s(x'''')))


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

SUM1(s(s(s(x'''')))) -> SUM1(s(s(x'''')))


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
SUM1(x1) -> SUM1(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


sum(0) -> 0
sum(s(x)) -> +(sum(x), s(x))
sum1(0) -> 0
sum1(s(x)) -> s(+(sum1(x), +(x, x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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