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
Polynomial Ordering
       →DP Problem 2
Polo
       →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 implicit AFS that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Dependency Graph
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polynomial 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))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  x1  
  POL(0)=  1  
  POL(minus(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →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
Polo
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polo
       →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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 11
Polynomial Ordering


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)))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(plus(x1, x2))=  x1 + x2  
  POL(PLUS(x1, x2))=  1 + x1 + x2  
  POL(0)=  0  
  POL(minus(x1, x2))=  1 + x1  
  POL(s(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 12
Polynomial Ordering


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(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)))





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(plus(x1, x2))=  1 + x1 + x2  
  POL(PLUS(x1, x2))=  1 + x1 + x2  
  POL(0)=  0  
  POL(minus(x1, x2))=  1 + x1  
  POL(s(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 13
Polynomial Ordering


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





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(plus(x1, x2))=  x1 + x2  
  POL(PLUS(x1, x2))=  1 + x1 + x2  
  POL(0)=  0  
  POL(minus(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 14
Dependency Graph


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


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.

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