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)

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

         ↳Remaining

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)

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)

The following rules can be oriented:

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)

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) -> x₁
top(x₁) -> top
check(x₁) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
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)
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)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
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)
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)

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