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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0)))

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →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)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0)))





The following dependency pair can be strictly oriented:

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


There are no usable rules 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:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


Rules:


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





The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


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

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


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

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


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


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0)))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 7
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(minus(x, s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x, s(0)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 8
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 9
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(plus(0, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(0))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 10
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(plus(s(x''), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(plus(x'', s(0))))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 11
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(minus(x, s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x, s(0)))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 12
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(minus(s(x'''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), minus(x''', 0))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 13
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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





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

PLUS(minus(s(x''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), minus(x'', 0))
two new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 14
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(minus(s(x'''), s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), x''')
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')


Rules:


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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 15
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')


Rules:


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





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

PLUS(plus(s(0), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(0)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 16
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')


Rules:


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





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

PLUS(plus(s(s(x')), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(plus(x', s(0)))))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 17
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(0))))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))


Rules:


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





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

PLUS(minus(x, s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 18
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(0)), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(0))))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))


Rules:


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





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

PLUS(minus(s(x''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), minus(x'', 0))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 19
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(0))))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))


Rules:


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





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

PLUS(minus(s(x'''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), minus(x''', 0))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 20
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(0))))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')


Rules:


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





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

PLUS(minus(s(x''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 21
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(0))))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')


Rules:


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





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

PLUS(plus(s(s(0)), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(0))))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 22
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')


Rules:


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





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

PLUS(plus(s(s(s(x''))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(plus(x'', s(0))))))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 23
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))


Rules:


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





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

PLUS(minus(x, s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x, s(0)))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 24
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))


Rules:


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





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

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 25
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))


Rules:


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





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

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 26
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')


Rules:


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





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

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), minus(x''', 0))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 27
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')


Rules:


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





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

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), minus(x'', 0))
three new Dependency Pairs are created:

PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 28
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')


Rules:


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





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

PLUS(plus(s(s(s(0))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(0)))))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 29
                 ↳Narrowing Transformation


Dependency Pairs:

PLUS(plus(s(s(s(0))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(s(0)))))
PLUS(plus(s(s(s(0))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(s(0)))))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')


Rules:


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





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

PLUS(plus(s(s(s(s(x')))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(plus(x', s(0)))))))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 30
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

PLUS(plus(s(s(s(s(s(x''))))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(s(plus(x'', s(0))))))))
PLUS(plus(s(s(s(s(0)))), s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), s(s(s(s(s(0))))))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(s(x')))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(s(plus(x', s(0)))))))
PLUS(plus(s(s(s(0))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(s(0)))))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x'))), s(s(s(y'))))) -> PLUS(minus(x', y'), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x0)))), s(s(s(s(y'')))))) -> PLUS(minus(x0, y''), minus(x'', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(s(x''))), s(s(s(y'))))) -> PLUS(minus(x'', y'), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(s(s(x')))), s(s(s(s(y'')))))) -> PLUS(minus(x', y''), minus(x''', 0))
PLUS(minus(s(x'''), s(0)), minus(s(s(s(x''''))), s(s(s(0))))) -> PLUS(x'''', minus(x''', 0))
PLUS(minus(s(x''), s(0)), minus(s(s(s(s(x''')))), s(s(s(s(y'')))))) -> PLUS(minus(x''', y''), minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(s(s(x''))))), s(s(s(s(s(y'))))))) -> PLUS(minus(x'', y'), minus(x, s(0)))
PLUS(minus(x, s(0)), minus(s(s(s(s(x'''')))), s(s(s(s(0)))))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(s(s(x''))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(s(x''))), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(plus(x'', s(0))))))
PLUS(plus(s(s(0)), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(0))))
PLUS(plus(s(s(0)), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(s(0))))
PLUS(minus(s(x'''), s(0)), minus(s(s(x0)), s(s(z'')))) -> PLUS(minus(x0, z''), x''')
PLUS(minus(s(x''), s(0)), minus(s(s(x0')), s(s(0)))) -> PLUS(x0', minus(x'', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x')), s(s(z'')))) -> PLUS(minus(x', z''), x'''')
PLUS(minus(s(x'''), s(0)), minus(s(s(x'')), s(s(0)))) -> PLUS(x'', minus(x''', 0))
PLUS(minus(s(x''''), s(0)), minus(s(s(x''')), s(s(z'')))) -> PLUS(minus(x''', z''), x'''')
PLUS(minus(s(x''), s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x'', 0))
PLUS(minus(x, s(0)), minus(s(s(s(x'''))), s(s(s(0))))) -> PLUS(x''', minus(x, s(0)))
PLUS(plus(s(s(x')), s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), s(s(plus(x', s(0)))))
PLUS(plus(s(s(x')), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(plus(x', s(0)))))
PLUS(plus(s(0), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(0)))
PLUS(plus(s(0), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(s(0)))
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x'''), s(0)), minus(s(x'), s(s(z')))) -> PLUS(minus(x', s(z')), x''')
PLUS(minus(s(x''''), s(0)), minus(s(x''), s(s(z')))) -> PLUS(minus(x'', s(z')), x'''')
PLUS(minus(x, s(0)), minus(s(s(x'''')), s(s(0)))) -> PLUS(x'''', minus(x, s(0)))
PLUS(plus(s(x''), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(plus(x'', s(0))))
PLUS(plus(s(x''), s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(plus(x'', s(0))))
PLUS(plus(0, s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(0))
PLUS(plus(0, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), s(0))
PLUS(plus(x, s(0)), plus(s(x''), s(s(z')))) -> PLUS(s(plus(x'', s(s(z')))), plus(x, s(0)))
PLUS(plus(x, s(0)), plus(0, s(s(z')))) -> PLUS(s(s(z')), plus(x, s(0)))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(plus(s(s(s(0))), s(0)), plus(s(x'), s(s(z')))) -> PLUS(s(plus(x', s(s(z')))), s(s(s(s(0)))))


Rules:


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




Termination of R could not be shown.
Duration:
0:13 minutes