Term Rewriting System R:
[a, x, k, d]
f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(a, empty) -> G(a, empty)
F(a, cons(x, k)) -> F(cons(x, a), k)
G(cons(x, k), d) -> G(k, cons(x, d))

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(cons(x, k), d) -> G(k, cons(x, d))


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




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

G(cons(x, k), d) -> G(k, cons(x, d))
one new Dependency Pair is created:

G(cons(x0, k''), cons(x'', d'')) -> G(k'', cons(x0, cons(x'', d'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(cons(x0, k''), cons(x'', d'')) -> G(k'', cons(x0, cons(x'', d'')))


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




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

G(cons(x0, k''), cons(x'', d'')) -> G(k'', cons(x0, cons(x'', d'')))
one new Dependency Pair is created:

G(cons(x0'', k''''), cons(x'''', cons(x''''', d''''))) -> G(k'''', cons(x0'', cons(x'''', cons(x''''', d''''))))

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

G(cons(x0'', k''''), cons(x'''', cons(x''''', d''''))) -> G(k'''', cons(x0'', cons(x'''', cons(x''''', d''''))))


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




The following dependency pair can be strictly oriented:

G(cons(x0'', k''''), cons(x'''', cons(x''''', d''''))) -> G(k'''', cons(x0'', cons(x'''', cons(x''''', d''''))))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(G(x1, x2))=  x1  
  POL(cons(x1, x2))=  1 + x2  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Instantiation Transformation


Dependency Pair:

F(a, cons(x, k)) -> F(cons(x, a), k)


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




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

F(a, cons(x, k)) -> F(cons(x, a), k)
one new Dependency Pair is created:

F(cons(x'', a''), cons(x0, k'')) -> F(cons(x0, cons(x'', a'')), k'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Instantiation Transformation


Dependency Pair:

F(cons(x'', a''), cons(x0, k'')) -> F(cons(x0, cons(x'', a'')), k'')


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




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

F(cons(x'', a''), cons(x0, k'')) -> F(cons(x0, cons(x'', a'')), k'')
one new Dependency Pair is created:

F(cons(x'''', cons(x''''', a'''')), cons(x0'', k'''')) -> F(cons(x0'', cons(x'''', cons(x''''', a''''))), k'''')

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

F(cons(x'''', cons(x''''', a'''')), cons(x0'', k'''')) -> F(cons(x0'', cons(x'''', cons(x''''', a''''))), k'''')


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(cons(x'''', cons(x''''', a'''')), cons(x0'', k'''')) -> F(cons(x0'', cons(x'''', cons(x''''', a''''))), k'''')


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x2  
  POL(F(x1, x2))=  x2  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


f(a, empty) -> g(a, empty)
f(a, cons(x, k)) -> f(cons(x, a), k)
g(empty, d) -> d
g(cons(x, k), d) -> g(k, cons(x, d))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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