Term Rewriting System R:
[x, y, z]
rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

REV(.(x, y)) -> ++'(rev(y), .(x, nil))
REV(.(x, y)) -> REV(y)
++'(.(x, y), z) -> ++'(y, z)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

++'(.(x, y), z) -> ++'(y, z)


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




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

++'(.(x, y), z) -> ++'(y, z)
one new Dependency Pair is created:

++'(.(x, .(x'', y'')), z'') -> ++'(.(x'', y''), z'')

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

++'(.(x, .(x'', y'')), z'') -> ++'(.(x'', y''), z'')


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




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

++'(.(x, .(x'', y'')), z'') -> ++'(.(x'', y''), z'')
one new Dependency Pair is created:

++'(.(x, .(x'''', .(x''''', y''''))), z'''') -> ++'(.(x'''', .(x''''', y'''')), z'''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

++'(.(x, .(x'''', .(x''''', y''''))), z'''') -> ++'(.(x'''', .(x''''', y'''')), z'''')


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

++'(.(x, .(x'''', .(x''''', y''''))), z'''') -> ++'(.(x'''', .(x''''', y'''')), z'''')


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(++'(x1, x2))=  1 + x1  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

REV(.(x, y)) -> REV(y)


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




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

REV(.(x, y)) -> REV(y)
one new Dependency Pair is created:

REV(.(x, .(x'', y''))) -> REV(.(x'', y''))

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 Pair:

REV(.(x, .(x'', y''))) -> REV(.(x'', y''))


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




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

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

REV(.(x, .(x'''', .(x''''', y'''')))) -> REV(.(x'''', .(x''''', y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pair:

REV(.(x, .(x'''', .(x''''', y'''')))) -> REV(.(x'''', .(x''''', y'''')))


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

REV(.(x, .(x'''', .(x''''', y'''')))) -> REV(.(x'''', .(x''''', y'''')))


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(REV(x1))=  1 + x1  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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