Term Rewriting System R:
[y, x]
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(s(x), y) -> +'(x, y)
+'(p(x), y) -> +'(x, y)
MINUS(s(x)) -> MINUS(x)
MINUS(p(x)) -> MINUS(x)
*'(s(x), y) -> +'(*(x, y), y)
*'(s(x), y) -> *'(x, y)
*'(p(x), y) -> +'(*(x, y), minus(y))
*'(p(x), y) -> *'(x, y)
*'(p(x), y) -> MINUS(y)

Furthermore, R contains three SCCs.


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


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(s(x), y) -> +'(x, y)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(p(x), y) -> +'(x, y)
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(s(s(x'')), y'') -> +'(s(x''), y'')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 6
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(s(p(x'')), y'') -> +'(p(x''), y'')
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(p(p(x'')), y'') -> +'(p(x''), y'')
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(p(s(s(x''''))), y') -> +'(s(s(x'''')), y')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 9
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(p(s(p(x''''))), y') -> +'(s(p(x'''')), y')
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 10
Argument Filtering and Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


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:
+'(x1, x2) -> +'(x1, x2)
p(x1) -> p(x1)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 11
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

MINUS(p(x)) -> MINUS(x)
MINUS(s(x)) -> MINUS(x)


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

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

MINUS(s(s(x''))) -> MINUS(s(x''))
MINUS(s(p(x''))) -> MINUS(p(x''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(s(p(x''))) -> MINUS(p(x''))
MINUS(s(s(x''))) -> MINUS(s(x''))
MINUS(p(x)) -> MINUS(x)


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

MINUS(p(x)) -> MINUS(x)
three new Dependency Pairs are created:

MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 13
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(x''))) -> MINUS(s(x''))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(s(p(x''))) -> MINUS(p(x''))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


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

MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 14
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(s(p(x''))) -> MINUS(p(x''))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

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

MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 15
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

MINUS(p(p(x''))) -> MINUS(p(x''))
three new Dependency Pairs are created:

MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 16
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
two new Dependency Pairs are created:

MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 17
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

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

MINUS(p(s(p(p(x''''''))))) -> MINUS(s(p(p(x''''''))))
MINUS(p(s(p(s(s(x'''''''')))))) -> MINUS(s(p(s(s(x'''''''')))))
MINUS(p(s(p(s(p(x'''''''')))))) -> MINUS(s(p(s(p(x'''''''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 18
Argument Filtering and Ordering
       →DP Problem 3
FwdInst


Dependency Pairs:

MINUS(p(s(p(s(p(x'''''''')))))) -> MINUS(s(p(s(p(x'''''''')))))
MINUS(p(s(p(s(s(x'''''''')))))) -> MINUS(s(p(s(s(x'''''''')))))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(p(x''''''))))) -> MINUS(s(p(p(x''''''))))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MINUS(p(s(p(s(p(x'''''''')))))) -> MINUS(s(p(s(p(x'''''''')))))
MINUS(p(s(p(s(s(x'''''''')))))) -> MINUS(s(p(s(s(x'''''''')))))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(p(x''''''))))) -> MINUS(s(p(p(x''''''))))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(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:
MINUS(x1) -> MINUS(x1)
s(x1) -> s(x1)
p(x1) -> p(x1)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 12
FwdInst
             ...
               →DP Problem 19
Dependency Graph
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(s(x), y) -> *'(x, y)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(p(x), y) -> *'(x, y)
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 21
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(s(s(x'')), y'') -> *'(s(x''), y'')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 22
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(s(p(x'')), y'') -> *'(p(x''), y'')
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 23
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(p(p(x'')), y'') -> *'(p(x''), y'')
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 24
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(p(s(s(x''''))), y') -> *'(s(s(x'''')), y')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 25
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(p(s(p(x''''))), y') -> *'(s(p(x'''')), y')
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 26
Argument Filtering and Ordering


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


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:
*'(x1, x2) -> *'(x1, x2)
p(x1) -> p(x1)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 20
FwdInst
             ...
               →DP Problem 27
Dependency Graph


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:04 minutes