Term Rewriting System R:
[x, y]
-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

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


Rule:


-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))


Strategy:

innermost




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

-'(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -'(x, y)
one new Dependency Pair is created:

-'(-(neg(-(neg(x'0), neg(x'''))), neg(-(neg(x'0), neg(x''')))), -(neg(-(neg(y'''), neg(y''''))), neg(-(neg(y'''), neg(y''''))))) -> -'(-(neg(x'0), neg(x''')), -(neg(y'''), neg(y'''')))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

-'(-(neg(-(neg(x'0), neg(x'''))), neg(-(neg(x'0), neg(x''')))), -(neg(-(neg(y'''), neg(y''''))), neg(-(neg(y'''), neg(y''''))))) -> -'(-(neg(x'0), neg(x''')), -(neg(y'''), neg(y'''')))
-'(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -'(-(x, y), -(x, y))


Rule:


-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

-'(-(neg(-(neg(x'0), neg(x'''))), neg(-(neg(x'0), neg(x''')))), -(neg(-(neg(y'''), neg(y''''))), neg(-(neg(y'''), neg(y''''))))) -> -'(-(neg(x'0), neg(x''')), -(neg(y'''), neg(y'''')))


Additionally, the following usable rule for innermost can be oriented:

-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  1 + x1  
  POL(neg(x1))=  x1  
  POL(-(x1, x2))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pair:

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


Rule:


-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rule for innermost can be oriented:

-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  x1  
  POL(neg(x1))=  1 + x1  
  POL(-(x1, x2))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rule:


-(-(neg(x), neg(x)), -(neg(y), neg(y))) -> -(-(x, y), -(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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