Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
F(s(x)) -> -'(s(x), g(f(x)))
F(s(x)) -> G(f(x))
F(s(x)) -> F(x)
G(s(x)) -> -'(s(x), f(g(x)))
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

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

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

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
         ↳Nar


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS 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:
-'(x1, x2) -> -'(x1, x2)
s(x1) -> s(x1)


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


Dependency Pair:


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(x)) -> G(f(x))
two new Dependency Pairs are created:

F(s(0)) -> G(0)
F(s(s(x''))) -> G(-(s(x''), g(f(x''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Narrowing Transformation


Dependency Pairs:

F(s(s(x''))) -> G(-(s(x''), g(f(x''))))
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(x)) -> F(g(x))
two new Dependency Pairs are created:

G(s(0)) -> F(s(0))
G(s(s(x''))) -> F(-(s(x''), f(g(x''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 7
                 ↳Narrowing Transformation


Dependency Pairs:

G(s(s(x''))) -> F(-(s(x''), f(g(x''))))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(x''))) -> G(-(s(x''), g(f(x''))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(s(x''))) -> G(-(s(x''), g(f(x''))))
two new Dependency Pairs are created:

F(s(s(0))) -> G(-(s(0), g(0)))
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 8
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(0))) -> G(-(s(0), g(0)))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(-(s(x''), f(g(x''))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(s(0))) -> G(-(s(0), g(0)))
one new Dependency Pair is created:

F(s(s(0))) -> G(-(s(0), s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 9
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(x''))) -> F(-(s(x''), f(g(x''))))
F(s(s(0))) -> G(-(s(0), s(0)))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(s(0))) -> G(-(s(0), s(0)))
one new Dependency Pair is created:

F(s(s(0))) -> G(-(0, 0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 10
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(s(0))) -> G(-(0, 0))
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(-(s(x''), f(g(x''))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(s(0))) -> G(-(0, 0))
one new Dependency Pair is created:

F(s(s(0))) -> G(0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 11
                 ↳Narrowing Transformation


Dependency Pairs:

G(s(s(x''))) -> F(-(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(x''))) -> F(-(s(x''), f(g(x''))))
two new Dependency Pairs are created:

G(s(s(0))) -> F(-(s(0), f(s(0))))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 12
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
G(s(s(0))) -> F(-(s(0), f(s(0))))
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(0))) -> F(-(s(0), f(s(0))))
one new Dependency Pair is created:

G(s(s(0))) -> F(-(s(0), -(s(0), g(f(0)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 13
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(0))) -> F(-(s(0), -(s(0), g(f(0)))))
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(0))) -> F(-(s(0), -(s(0), g(f(0)))))
one new Dependency Pair is created:

G(s(s(0))) -> F(-(s(0), -(s(0), g(0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 14
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(0))) -> F(-(s(0), -(s(0), g(0))))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(0))) -> F(-(s(0), -(s(0), g(0))))
one new Dependency Pair is created:

G(s(s(0))) -> F(-(s(0), -(s(0), s(0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 15
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(0))) -> F(-(s(0), -(s(0), s(0))))
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(0))) -> F(-(s(0), -(s(0), s(0))))
one new Dependency Pair is created:

G(s(s(0))) -> F(-(s(0), -(0, 0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 16
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(0))) -> F(-(s(0), -(0, 0)))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(0))) -> F(-(s(0), -(0, 0)))
one new Dependency Pair is created:

G(s(s(0))) -> F(-(s(0), 0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 17
                 ↳Rewriting Transformation


Dependency Pairs:

G(s(s(0))) -> F(-(s(0), 0))
G(s(0)) -> F(s(0))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(0))) -> F(-(s(0), 0))
one new Dependency Pair is created:

G(s(s(0))) -> F(s(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 18
                 ↳Forward Instantiation Transformation


Dependency Pairs:

G(s(s(0))) -> F(s(0))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(0)) -> F(s(0))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(x)) -> F(x)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 19
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
F(s(s(x''))) -> F(s(x''))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(x)) -> G(x)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 20
                 ↳Forward Instantiation Transformation


Dependency Pairs:

G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))
G(s(s(x''))) -> G(s(x''))
F(s(s(x''))) -> F(s(x''))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

F(s(s(x''))) -> F(s(x''))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 21
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(x''))) -> G(s(x''))
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




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

G(s(s(x''))) -> G(s(x''))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 22
                 ↳Argument Filtering and Ordering


Dependency Pairs:

G(s(s(s(s(s(x''''')))))) -> G(s(s(s(s(x''''')))))
G(s(s(s(x'''')))) -> G(s(s(x'''')))
G(s(s(s(s(x''''))))) -> G(s(s(s(x''''))))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

G(s(s(s(s(s(x''''')))))) -> G(s(s(s(s(x''''')))))
G(s(s(s(x'''')))) -> G(s(s(x'''')))
G(s(s(s(s(x''''))))) -> G(s(s(s(x''''))))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(s(x')))) -> F(-(s(s(x')), f(-(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(-(s(s(x')), g(-(s(x'), g(f(x'))))))
F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))


The following usable rules for innermost w.r.t. to the AFS can be oriented:

-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{F, s} > G

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
s(x1) -> s(x1)
G(x1) -> G(x1)
-(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 23
                 ↳Dependency Graph


Dependency Pair:


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes