Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

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)
DOUBLE(s(x)) -> DOUBLE(x)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> MINUS(x, y)
PLUS(s(x), y) -> DOUBLE(y)
PLUS(s(plus(x, y)), z) -> PLUS(plus(x, y), z)

Furthermore, R contains three SCCs.


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


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


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 4
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
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)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


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 4
FwdInst
             ...
               →DP Problem 5
Argument Filtering and Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
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)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


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
Nar


Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

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

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

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

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 8
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pair:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(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:
DOUBLE(x1) -> DOUBLE(x1)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


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
Narrowing Transformation


Dependency Pairs:

PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
four new Dependency Pairs are created:

PLUS(s(x''), 0) -> PLUS(x'', double(0))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))
PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rewriting Transformation


Dependency Pairs:

PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))
PLUS(s(x''), 0) -> PLUS(x'', double(0))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(x, s(y))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(x''), 0) -> PLUS(x'', double(0))
one new Dependency Pair is created:

PLUS(s(x''), 0) -> PLUS(x'', 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 11
Rewriting Transformation


Dependency Pairs:

PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

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

PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 12
Rewriting Transformation


Dependency Pairs:

PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x''), 0) -> PLUS(x'', 0)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 13
Narrowing Transformation


Dependency Pairs:

PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
three new Dependency Pairs are created:

PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 14
Forward Instantiation Transformation


Dependency Pairs:

PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), 0) -> PLUS(x, 0)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(x), y) -> PLUS(x, y)
seven new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 15
Forward Instantiation Transformation


Dependency Pairs:

PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x''), 0) -> PLUS(x'', 0)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(x), y) -> PLUS(x, s(y))
nine new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 16
Forward Instantiation Transformation


Dependency Pairs:

PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(x'')), y'') -> PLUS(s(x''), s(y''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
20 new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 17
Forward Instantiation Transformation


Dependency Pairs:

PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(x'')), y'') -> PLUS(s(x''), s(y''))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




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

PLUS(s(s(x'')), y'') -> PLUS(s(x''), s(y''))
21 new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 18
Argument Filtering and Ordering


Dependency Pairs:

PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), s(0))
PLUS(s(s(s(x'''''))), 0) -> PLUS(s(s(x''''')), s(0))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(s(s(s(x'0'''''''')))))), y''') -> PLUS(s(s(s(s(s(x'0''''''''))))), s(y'''))
PLUS(s(s(s(s(s(s(x'''''''''')))))), y''') -> PLUS(s(s(s(s(s(x''''''''''))))), s(y'''))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(s(x''''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y''') -> PLUS(s(s(s(s(x'0'''''')))), s(y'''))
PLUS(s(s(s(s(s(x''''''''))))), y''') -> PLUS(s(s(s(s(x'''''''')))), s(y'''))
PLUS(s(s(s(x'''''))), s(x''''''')) -> PLUS(s(s(x''''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), s(y''''))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), s(y''''))
PLUS(s(s(s(x''''))), y''') -> PLUS(s(s(x'''')), s(y'''))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), s(y''''))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y''') -> PLUS(s(s(s(x'0''''))), s(y'''))
PLUS(s(s(s(s(x'''''')))), y''') -> PLUS(s(s(s(x''''''))), s(y'''))
PLUS(s(s(x''')), s(x''''')) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y''') -> PLUS(s(s(x'0'')), s(y'''))
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), s(0))
PLUS(s(s(s(x'''''))), 0) -> PLUS(s(s(x''''')), s(0))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(s(s(s(x'0'''''''')))))), y''') -> PLUS(s(s(s(s(s(x'0''''''''))))), s(y'''))
PLUS(s(s(s(s(s(s(x'''''''''')))))), y''') -> PLUS(s(s(s(s(s(x''''''''''))))), s(y'''))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(s(x''''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y''') -> PLUS(s(s(s(s(x'0'''''')))), s(y'''))
PLUS(s(s(s(s(s(x''''''''))))), y''') -> PLUS(s(s(s(s(x'''''''')))), s(y'''))
PLUS(s(s(s(x'''''))), s(x''''''')) -> PLUS(s(s(x''''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), s(y''''))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), s(y''''))
PLUS(s(s(s(x''''))), y''') -> PLUS(s(s(x'''')), s(y'''))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), s(y''''))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y''') -> PLUS(s(s(s(x'0''''))), s(y'''))
PLUS(s(s(s(s(x'''''')))), y''') -> PLUS(s(s(s(x''''''))), s(y'''))
PLUS(s(s(x''')), s(x''''')) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y''') -> PLUS(s(s(x'0'')), s(y'''))
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))


The following usable rules for innermost can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
PLUS > double > s

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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Rw
             ...
               →DP Problem 19
Dependency Graph


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:48 minutes