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))

Innermost 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))


Strategy:

innermost




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))


Strategy:

innermost




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
                 ↳Argument Filtering and Ordering


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
R1 > cons

resulting in one new DP problem.
Used Argument Filtering System:
R1(x1, x2) -> R1(x1, x2)
cons(x1, x2) -> cons(x1, x2)


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


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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