Term Rewriting System R:
[ls, a, x, k]
rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

REV(ls) -> R1(ls, empty)
R1(cons(x, k), a) -> R1(k, cons(x, a))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Instantiation Transformation


Dependency Pair:

R1(cons(x, k), a) -> R1(k, cons(x, a))


Rules:


rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))





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

R1(cons(x, k), a) -> R1(k, cons(x, a))
one new Dependency Pair is created:

R1(cons(x0, k''), cons(x'', a'')) -> R1(k'', cons(x0, cons(x'', a'')))

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

R1(cons(x0, k''), cons(x'', a'')) -> R1(k'', cons(x0, cons(x'', a'')))


Rules:


rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))





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

R1(cons(x0, k''), cons(x'', a'')) -> R1(k'', cons(x0, cons(x'', a'')))
one new Dependency Pair is created:

R1(cons(x0'', k''''), cons(x'''', cons(x''''', a''''))) -> R1(k'''', cons(x0'', cons(x'''', 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 Pair:

R1(cons(x0'', k''''), cons(x'''', cons(x''''', a''''))) -> R1(k'''', cons(x0'', cons(x'''', cons(x''''', a''''))))


Rules:


rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))





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

R1(cons(x0'', k''''), cons(x'''', cons(x''''', a''''))) -> R1(k'''', cons(x0'', cons(x'''', cons(x''''', a''''))))
one new Dependency Pair is created:

R1(cons(x0'''', k''''''), cons(x''''0, cons(x'''''0, cons(x'''''''', a'''''')))) -> R1(k'''''', cons(x0'''', cons(x''''0, cons(x'''''0, cons(x'''''''', a'''''')))))

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 Pair:

R1(cons(x0'''', k''''''), cons(x''''0, cons(x'''''0, cons(x'''''''', a'''''')))) -> R1(k'''''', cons(x0'''', cons(x''''0, cons(x'''''0, cons(x'''''''', a'''''')))))


Rules:


rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))





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

R1(cons(x0'''', k''''''), cons(x''''0, cons(x'''''0, cons(x'''''''', a'''''')))) -> R1(k'''''', cons(x0'''', cons(x''''0, cons(x'''''0, cons(x'''''''', a'''''')))))
one new Dependency Pair is created:

R1(cons(x0'''''', k''''''''), cons(x''''0'', cons(x'''''0'', cons(x'''''''''', cons(x''''''''''', a''''''''))))) -> R1(k'''''''', cons(x0'''''', cons(x''''0'', cons(x'''''0'', cons(x'''''''''', 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
                 ↳Instantiation Transformation


Dependency Pair:

R1(cons(x0'''''', k''''''''), cons(x''''0'', cons(x'''''0'', cons(x'''''''''', cons(x''''''''''', a''''''''))))) -> R1(k'''''''', cons(x0'''''', cons(x''''0'', cons(x'''''0'', cons(x'''''''''', cons(x''''''''''', a''''''''))))))


Rules:


rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))





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

R1(cons(x0'''''', k''''''''), cons(x''''0'', cons(x'''''0'', cons(x'''''''''', cons(x''''''''''', a''''''''))))) -> R1(k'''''''', cons(x0'''''', cons(x''''0'', cons(x'''''0'', cons(x'''''''''', cons(x''''''''''', a''''''''))))))
one new Dependency Pair is created:

R1(cons(x0'''''''', k''''''''''), cons(x''''0'''', cons(x'''''0'''', cons(x''''''''''0, cons(x'''''''''''0, cons(x'''''''''''''', a'''''''''')))))) -> R1(k'''''''''', cons(x0'''''''', cons(x''''0'''', cons(x'''''0'''', cons(x''''''''''0, cons(x'''''''''''0, cons(x'''''''''''''', a'''''''''')))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 6
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

R1(cons(x0'''''''', k''''''''''), cons(x''''0'''', cons(x'''''0'''', cons(x''''''''''0, cons(x'''''''''''0, cons(x'''''''''''''', a'''''''''')))))) -> R1(k'''''''''', cons(x0'''''''', cons(x''''0'''', cons(x'''''0'''', cons(x''''''''''0, cons(x'''''''''''0, cons(x'''''''''''''', a'''''''''')))))))


Rules:


rev(ls) -> r1(ls, empty)
r1(empty, a) -> a
r1(cons(x, k), a) -> r1(k, cons(x, a))




Termination of R could not be shown.
Duration:
0:00 minutes