Termination Proof using AProVE

Term Rewriting System R:
[X, X1, X2]
active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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(f(0)) -> CONS(0, f(s(0)))
ACTIVE(f(0)) -> F(s(0))
ACTIVE(f(0)) -> S(0)
ACTIVE(f(s(0))) -> F(p(s(0)))
ACTIVE(f(s(0))) -> P(s(0))
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(p(X)) -> P(active(X))
ACTIVE(p(X)) -> ACTIVE(X)
F(mark(X)) -> F(X)
F(ok(X)) -> F(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
P(mark(X)) -> P(X)
P(ok(X)) -> P(X)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(p(X)) -> P(proper(X))
PROPER(p(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 seven SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁) -> F(x₁)
mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

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

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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 2

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

S(x₁) -> S(x₁)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

P(ok(X)) -> P(X)
P(mark(X)) -> P(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

P(ok(X)) -> P(X)
P(mark(X)) -> P(X)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

P(x₁) -> P(x₁)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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

         ↳Argument Filtering and Ordering

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

ACTIVE(x₁) -> ACTIVE(x₁)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
p(x₁) -> p(x₁)
f(x₁) -> f(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 12

             ↳Dependency Graph

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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

         ↳AFS

       →DP Problem 6

         ↳Argument Filtering and Ordering

       →DP Problem 7

         ↳Remaining

Dependency Pairs:

PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

PROPER(x₁) -> PROPER(x₁)
f(x₁) -> f(x₁)
p(x₁) -> p(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

           →DP Problem 13

             ↳Dependency Graph

       →DP Problem 7

         ↳Remaining

Dependency Pair:

Rules:

active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳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(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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:
0:23 minutes