Termination Proof using AProVE

Term Rewriting System R:
[x, y]
top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

TOP(sent(x)) -> TOP(check(rest(x)))
TOP(sent(x)) -> CHECK(rest(x))
TOP(sent(x)) -> REST(x)
CHECK(sent(x)) -> CHECK(x)
CHECK(rest(x)) -> REST(check(x))
CHECK(rest(x)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

CHECK(x₁) -> CHECK(x₁)
rest(x₁) -> rest(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
sent(x₁) -> sent(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pair:

TOP(sent(x)) -> TOP(check(rest(x)))

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Strategy:

innermost

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

TOP(sent(x)) -> TOP(check(rest(x)))

three new Dependency Pairs are created:

TOP(sent(x'')) -> TOP(rest(check(x'')))
TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(cons(x'', y'))) -> TOP(check(sent(y')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Argument Filtering and Ordering

Dependency Pairs:

TOP(sent(cons(x'', y'))) -> TOP(check(sent(y')))
TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(x'')) -> TOP(rest(check(x'')))

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Strategy:

innermost

The following dependency pair can be strictly oriented:

TOP(sent(cons(x'', y'))) -> TOP(check(sent(y')))

The following usable rules for innermost can be oriented:

check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)

Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

TOP(x₁) -> TOP(x₁)
sent(x₁) -> x₁
cons(x₁, x₂) -> cons(x₁, x₂)
check(x₁) -> x₁
rest(x₁) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳AFS

...

               →DP Problem 5

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(x'')) -> TOP(rest(check(x'')))

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Strategy:

innermost

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