Term Rewriting System R:
[m, n, r]
p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

P(m, n, s(r)) -> P(m, r, n)
P(m, s(n), 0) -> P(0, n, m)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

P(m, s(n), 0) -> P(0, n, m)
P(m, n, s(r)) -> P(m, r, n)


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




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

P(m, n, s(r)) -> P(m, r, n)
two new Dependency Pairs are created:

P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)
P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
P(m, s(n), 0) -> P(0, n, m)


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




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

P(m, s(n), 0) -> P(0, n, m)
three new Dependency Pairs are created:

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))

The transformation is resulting in two new DP problems: 



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


Dependency Pair:

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




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

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)
one new Dependency Pair is created:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pair:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




The following dependency pair can be strictly oriented:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(P(x1, x2, x3))=  1 + x2  
  POL(0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳Dependency Graph


Dependency Pair:


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))
P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




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

P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
two new Dependency Pairs are created:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))
P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)
P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




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

P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)
two new Dependency Pairs are created:

P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(s(s(n'''''')), 0, s(s(0))) -> P(s(s(n'''''')), s(0), 0)

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




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

P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
two new Dependency Pairs are created:

P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))
P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Polynomial Ordering


Dependency Pairs:

P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))
P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))
P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))
P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




The following dependency pairs can be strictly oriented:

P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))
P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(P(x1, x2, x3))=  x1  
  POL(0)=  0  
  POL(s(x1))=  1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳Dependency Graph


Dependency Pairs:

P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))
P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 11
                 ↳Polynomial Ordering


Dependency Pair:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




The following dependency pair can be strictly oriented:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(P(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes