Termination Proof using AProVE

Term Rewriting System R:
[X, XS, X1, X2]
active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(zeros) -> CONS(0, zeros)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(tail(X)) -> TAIL(active(X))
ACTIVE(tail(X)) -> ACTIVE(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
TAIL(mark(X)) -> TAIL(X)
TAIL(ok(X)) -> TAIL(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(tail(X)) -> TAIL(proper(X))
PROPER(tail(X)) -> PROPER(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

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

CONS(x₁, x₂) -> CONS(x₁, x₂)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

TAIL(ok(X)) -> TAIL(X)
TAIL(mark(X)) -> TAIL(X)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

TAIL(ok(X)) -> TAIL(X)
TAIL(mark(X)) -> TAIL(X)

TAIL(x₁) -> TAIL(x₁)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

ACTIVE(tail(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

ACTIVE(tail(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

ACTIVE(x₁) -> ACTIVE(x₁)
tail(x₁) -> tail(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

PROPER(tail(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PROPER(tail(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

PROPER(x₁) -> PROPER(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
tail(x₁) -> tail(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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