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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) -> P(-(s(x), y))
F(s(x), y) -> -'(s(x), y)
F(s(x), y) -> P(-(y, s(x)))
F(s(x), y) -> -'(y, s(x))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) -> P(-(x, s(y)))
F(x, s(y)) -> -'(x, s(y))
F(x, s(y)) -> P(-(s(y), x))
F(x, s(y)) -> -'(s(y), x)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
-'(x1, x2) -> -'(x1, x2)
s(x1) -> s(x1)


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


Dependency Pair:


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

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

F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rewriting Transformation


Dependency Pairs:

F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

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

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 5
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 6
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
one new Dependency Pair is created:

F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 7
                 ↳Narrowing Transformation


Dependency Pairs:

F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

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

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 8
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 9
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
one new Dependency Pair is created:

F(0, s(y')) -> F(p(-(0, s(y'))), y')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 10
                 ↳Rewriting Transformation


Dependency Pairs:

F(0, s(y')) -> F(p(-(0, s(y'))), y')
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
one new Dependency Pair is created:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 11
                 ↳Narrowing Transformation


Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(0, s(y')) -> F(p(-(0, s(y'))), y')


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 12
                 ↳Narrowing Transformation


Dependency Pairs:

F(0, s(y')) -> F(p(-(0, s(y'))), y')
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




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

F(0, s(y')) -> F(p(-(0, s(y'))), y')
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 13
                 ↳Argument Filtering and Ordering


Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))


The following usable rules for innermost can be oriented:

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


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2) -> F(x1, x2)
s(x1) -> s(x1)
p(x1) -> x1
-(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 14
                 ↳Dependency Graph


Dependency Pair:


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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