Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2, X3]
active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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(X, g(X), Y)) -> F(Y, Y, Y)
ACTIVE(g(X)) -> G(active(X))
ACTIVE(g(X)) -> ACTIVE(X)
G(mark(X)) -> G(X)
G(ok(X)) -> G(X)
PROPER(f(X1, X2, X3)) -> F(proper(X1), proper(X2), proper(X3))
PROPER(f(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(g(X)) -> G(proper(X))
PROPER(g(X)) -> PROPER(X)
F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)
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

         ↳Nar

Dependency Pair:

F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)

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, x₂, x₃) -> F(x₁, x₂, x₃)
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

         ↳Nar

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Nar

Dependency Pairs:

G(ok(X)) -> G(X)
G(mark(X)) -> G(X)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

G(ok(X)) -> G(X)
G(mark(X)) -> G(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:

G(x₁) -> G(x₁)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(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

         ↳Nar

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Nar

Dependency Pair:

ACTIVE(g(X)) -> ACTIVE(X)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVE(g(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)
g(x₁) -> g(x₁)

     ↳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

         ↳Nar

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Nar

Dependency Pairs:

PROPER(g(X)) -> PROPER(X)
PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X1)

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PROPER(g(X)) -> PROPER(X)
PROPER(f(X1, X2, X3)) -> PROPER(X3)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X1)

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, x₂, x₃) -> f(x₁, x₂, x₃)
g(x₁) -> g(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

         ↳Nar

Dependency Pair:

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
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

         ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

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

four new Dependency Pairs are created:

TOP(mark(f(X1', X2', X3'))) -> TOP(f(proper(X1'), proper(X2'), proper(X3')))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(mark(b)) -> TOP(ok(b))
TOP(mark(c)) -> TOP(ok(c))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pairs:

TOP(mark(c)) -> TOP(ok(c))
TOP(mark(b)) -> TOP(ok(b))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(mark(f(X1', X2', X3'))) -> TOP(f(proper(X1'), proper(X2'), proper(X3')))
TOP(ok(X)) -> TOP(active(X))

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

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

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

four new Dependency Pairs are created:

TOP(ok(f(X'', g(X''), Y'))) -> TOP(mark(f(Y', Y', Y')))
TOP(ok(g(b))) -> TOP(mark(c))
TOP(ok(b)) -> TOP(mark(c))
TOP(ok(g(X''))) -> TOP(g(active(X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

TOP(ok(g(X''))) -> TOP(g(active(X'')))
TOP(ok(f(X'', g(X''), Y'))) -> TOP(mark(f(Y', Y', Y')))
TOP(mark(f(X1', X2', X3'))) -> TOP(f(proper(X1'), proper(X2'), proper(X3')))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))

Rules:

active(f(X, g(X), Y)) -> mark(f(Y, Y, Y))
active(g(b)) -> mark(c)
active(b) -> mark(c)
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
g(ok(X)) -> ok(g(X))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(g(X)) -> g(proper(X))
proper(b) -> ok(b)
proper(c) -> ok(c)
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:15 minutes