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
Polynomial 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 w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(cons(x1, x2)) =  1 + x2 POL(R1(x1, x2)) =  x1

resulting in one new DP problem.

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