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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

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


Rules:


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

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

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

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


Rules:


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

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

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(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
Polynomial Ordering
       →DP Problem 2
Nar


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.



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


Dependency Pair:


Rules:


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


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


Rules:


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

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(x''), g(f(x''))))
two new Dependency Pairs are created:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), g(s(0))))
one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(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 9
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), minus(s(0), f(g(0)))))
one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(s(0), 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 10
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), minus(s(0), f(0))))
one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(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 11
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), minus(s(0), s(0))))
one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(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 12
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), minus(0, 0)))
one new Dependency Pair is created:

F(s(s(0))) -> G(minus(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 13
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), 0))
one new Dependency Pair is created:

F(s(s(0))) -> G(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 14
Narrowing Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(x''), f(g(x''))))
two new Dependency Pairs are created:

G(s(s(0))) -> F(minus(s(0), f(0)))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(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 15
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), f(0)))
one new Dependency Pair is created:

G(s(s(0))) -> F(minus(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 16
Rewriting Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(s(0), s(0)))
one new Dependency Pair is created:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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(minus(0, 0))
one new Dependency Pair is created:

G(s(s(0))) -> 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 18
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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)
four new Dependency Pairs are created:

F(s(s(x''))) -> F(s(x''))
F(s(s(0))) -> F(s(0))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
F(s(s(s(0)))) -> F(s(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 19
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(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''))
four 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''''')))))
F(s(s(s(s(0))))) -> F(s(s(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 21
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


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


Rules:


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


Additionally, the following usable rules for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(g(x1))=  x1  
  POL(G(x1))=  x1  
  POL(minus(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(f(x1))=  1 + x1  
  POL(F(x1))=  x1  

resulting in one new DP problem.



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


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes