Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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(b, c, x)) -> F(x, x, x)
ACTIVE(f(x, y, z)) -> F(x, y, active(z))
ACTIVE(f(x, y, z)) -> ACTIVE(z)
F(x, y, mark(z)) -> F(x, y, z)
F(ok(x), ok(y), ok(z)) -> F(x, y, z)
PROPER(f(x, y, z)) -> F(proper(x), proper(y), proper(z))
PROPER(f(x, y, z)) -> PROPER(x)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(z)
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 four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

F(x, y, mark(z)) -> F(x, y, z)
F(ok(x), ok(y), ok(z)) -> F(x, y, z)

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(x, y, mark(z)) -> F(x, y, z)
F(ok(x), ok(y), ok(z)) -> F(x, y, z)

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:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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

         ↳Remaining

Dependency Pair:

ACTIVE(f(x, y, z)) -> ACTIVE(z)

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVE(f(x, y, z)) -> ACTIVE(z)

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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

         ↳Remaining

Dependency Pairs:

PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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

         ↳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(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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