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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(cons(x, k), l) -> G(k, l, cons(x, k))
G(a, b, c) -> F(a, cons(b, c))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Instantiation Transformation


Dependency Pairs:

G(a, b, c) -> F(a, cons(b, c))
F(cons(x, k), l) -> G(k, l, cons(x, k))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))


Strategy:

innermost




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

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

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))
G(a, b, c) -> F(a, cons(b, c))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))


Strategy:

innermost




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

G(a, b, c) -> F(a, cons(b, c))
one new Dependency Pair is created:

G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))
F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))


Strategy:

innermost




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

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))
one new Dependency Pair is created:

F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))
G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))


Strategy:

innermost




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

G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))
one new Dependency Pair is created:

G(a'', cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')) -> F(a'', cons(cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

G(a'', cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')) -> F(a'', cons(cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')))
F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))


Strategy:

innermost




The following dependency pair can be strictly oriented:

G(a'', cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')) -> F(a'', cons(cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')))


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(G(x1, x2, x3))=  1 + x1  
  POL(cons(x1, x2))=  1 + x2  
  POL(F(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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