Termination w.r.t. Q proof of AProVE_04_JFP

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔, 0 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPSizeChangeProof (⇔, 0 ms)

        ↳9 YES

    ↳10 QDP

        ↳11 UsableRulesProof (⇔, 0 ms)

        ↳12 QDP

        ↳13 QDPSizeChangeProof (⇔, 0 ms)

        ↳14 YES

    ↳15 QDP

        ↳16 UsableRulesProof (⇔, 0 ms)

        ↳17 QDP

        ↳18 QDPSizeChangeProof (⇔, 0 ms)

        ↳19 YES

    ↳20 QDP

        ↳21 UsableRulesReductionPairsProof (⇔, 0 ms)

        ↳22 QDP

        ↳23 TransformationProof (⇔, 0 ms)

        ↳24 QDP

        ↳25 DependencyGraphProof (⇔, 0 ms)

        ↳26 QDP

        ↳27 TransformationProof (⇔, 0 ms)

        ↳28 QDP

        ↳29 DependencyGraphProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 SemLabProof (⇒, 9249 ms)

        ↳32 QDP

        ↳33 DependencyGraphProof (⇔, 0 ms)

        ↳34 QDP

        ↳35 QDPOrderProof (⇔, 0 ms)

        ↳36 QDP

        ↳37 QDPOrderProof (⇔, 15 ms)

        ↳38 QDP

        ↳39 QDPOrderProof (⇔, 0 ms)

        ↳40 QDP

        ↳41 QDPOrderProof (⇔, 0 ms)

        ↳42 QDP

        ↳43 QDPOrderProof (⇔, 4 ms)

        ↳44 QDP

        ↳45 QDPOrderProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 QDPOrderProof (⇔, 17 ms)

        ↳48 QDP

        ↳49 QDPOrderProof (⇔, 0 ms)

        ↳50 QDP

        ↳51 PisEmptyProof (⇔, 0 ms)

        ↳52 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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)
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)
F(ok(x), ok(y), ok(z)) → F(x, y, z)
TOP(mark(x)) → TOP(proper(x))
TOP(mark(x)) → PROPER(x)
TOP(ok(x)) → TOP(active(x))
TOP(ok(x)) → ACTIVE(x)

The TRS R consists of the following 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)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(ok(x), ok(y), ok(z)) → F(x, y, z)
F(x, y, mark(z)) → F(x, y, z)

The TRS R consists of the following 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)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(ok(x), ok(y), ok(z)) → F(x, y, z)
F(x, y, mark(z)) → F(x, y, z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

F(ok(x), ok(y), ok(z)) → F(x, y, z)
The graph contains the following edges 1 > 1, 2 > 2, 3 > 3
F(x, y, mark(z)) → F(x, y, z)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

(9) YES

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(z)

The TRS R consists of the following 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)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) UsableRulesProof (EQUIVALENT transformation)

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

PROPER(f(x, y, z)) → PROPER(y)
The graph contains the following edges 1 > 1
PROPER(f(x, y, z)) → PROPER(x)
The graph contains the following edges 1 > 1
PROPER(f(x, y, z)) → PROPER(z)
The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(f(x, y, z)) → ACTIVE(z)

The TRS R consists of the following 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)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) UsableRulesProof (EQUIVALENT transformation)

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(f(x, y, z)) → ACTIVE(z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

ACTIVE(f(x, y, z)) → ACTIVE(z)
The graph contains the following edges 1 > 1

(19) YES

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(x)) → TOP(active(x))
TOP(mark(x)) → TOP(proper(x))

The TRS R consists of the following 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)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

top(mark(x)) → top(proper(x))
top(ok(x)) → top(active(x))

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(TOP(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(b) = 0
POL(c) = 0
POL(d) = 0
POL(f(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(m(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(x)) → TOP(active(x))
TOP(mark(x)) → TOP(proper(x))

The TRS R consists of the following rules:

proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule TOP(ok(x)) → TOP(active(x)) at position [0] we obtained the following new rules [LPAR04]:

TOP(ok(f(b, c, x0))) → TOP(mark(f(x0, x0, x0))) → TOP(ok(f(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2))) → TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))
TOP(ok(d)) → TOP(m(b)) → TOP(ok(d)) → TOP(m(b))
TOP(ok(d)) → TOP(mark(c)) → TOP(ok(d)) → TOP(mark(c))

(24) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(mark(x)) → TOP(proper(x))
TOP(ok(f(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))
TOP(ok(d)) → TOP(m(b))
TOP(ok(d)) → TOP(mark(c))

The TRS R consists of the following rules:

proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(26) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(mark(x)) → TOP(proper(x))
TOP(ok(f(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))
TOP(ok(d)) → TOP(mark(c))

The TRS R consists of the following rules:

proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(27) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule TOP(mark(x)) → TOP(proper(x)) at position [0] we obtained the following new rules [LPAR04]:

TOP(mark(b)) → TOP(ok(b)) → TOP(mark(b)) → TOP(ok(b))
TOP(mark(c)) → TOP(ok(c)) → TOP(mark(c)) → TOP(ok(c))
TOP(mark(d)) → TOP(ok(d)) → TOP(mark(d)) → TOP(ok(d))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2))) → TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))

(28) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(f(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))
TOP(ok(d)) → TOP(mark(c))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(c)) → TOP(ok(c))
TOP(mark(d)) → TOP(ok(d))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))

The TRS R consists of the following rules:

proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(29) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(f(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))

The TRS R consists of the following rules:

proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(31) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1.
active: 0
b: 1
c: 0
d: 0
proper: x₀
f: 0
m: 0
TOP: 0
ok: x₀
mark: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-0(b., c., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(ok.0(f.1-0-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.1(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(33) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(ok.0(f.1-0-0(b., c., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = x₁
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(d.) = 1
POL(f.0-0-0(x₁, x₂, x₃)) = 1
POL(f.0-0-1(x₁, x₂, x₃)) = 1
POL(f.0-1-0(x₁, x₂, x₃)) = 1
POL(f.0-1-1(x₁, x₂, x₃)) = 1
POL(f.1-0-0(x₁, x₂, x₃)) = 1
POL(f.1-0-1(x₁, x₂, x₃)) = 1 + x₃
POL(f.1-1-0(x₁, x₂, x₃)) = x₁ + x₃
POL(f.1-1-1(x₁, x₂, x₃)) = x₁
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = 1
POL(mark.1(x₁)) = 1 + x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = 1
POL(proper.1(x₁)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
proper.1(b.) → ok.1(b.)
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(ok.0(f.1-0-0(b., c., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(ok.0(f.1-0-0(b., c., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = 1
POL(active.1(x₁)) = 0
POL(b.) = 1
POL(c.) = 1
POL(d.) = 0
POL(f.0-0-0(x₁, x₂, x₃)) = 0
POL(f.0-0-1(x₁, x₂, x₃)) = 0
POL(f.0-1-0(x₁, x₂, x₃)) = 0
POL(f.0-1-1(x₁, x₂, x₃)) = 0
POL(f.1-0-0(x₁, x₂, x₃)) = x₁
POL(f.1-0-1(x₁, x₂, x₃)) = 0
POL(f.1-1-0(x₁, x₂, x₃)) = 0
POL(f.1-1-1(x₁, x₂, x₃)) = 0
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = x₁
POL(mark.1(x₁)) = x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = 0
POL(proper.1(x₁)) = x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
proper.1(b.) → ok.1(b.)
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))

(38) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(ok.0(f.1-0-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 1
POL(c.) = 0
POL(d.) = 0
POL(f.0-0-0(x₁, x₂, x₃)) = 0
POL(f.0-0-1(x₁, x₂, x₃)) = 0
POL(f.0-1-0(x₁, x₂, x₃)) = 1
POL(f.0-1-1(x₁, x₂, x₃)) = 0
POL(f.1-0-0(x₁, x₂, x₃)) = 1 + x₁
POL(f.1-0-1(x₁, x₂, x₃)) = 1 + x₁
POL(f.1-1-0(x₁, x₂, x₃)) = 1
POL(f.1-1-1(x₁, x₂, x₃)) = 1
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = x₁
POL(mark.1(x₁)) = x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = 0
POL(proper.1(x₁)) = x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
proper.1(b.) → ok.1(b.)
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))

(40) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(d.) = 0
POL(f.0-0-0(x₁, x₂, x₃)) = 1
POL(f.0-0-1(x₁, x₂, x₃)) = x₃
POL(f.0-1-0(x₁, x₂, x₃)) = 1
POL(f.0-1-1(x₁, x₂, x₃)) = 1
POL(f.1-0-0(x₁, x₂, x₃)) = 1
POL(f.1-0-1(x₁, x₂, x₃)) = 1
POL(f.1-1-0(x₁, x₂, x₃)) = 1
POL(f.1-1-1(x₁, x₂, x₃)) = 1
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = 1
POL(mark.1(x₁)) = x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = 0
POL(proper.1(x₁)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
proper.1(b.) → ok.1(b.)
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))

(42) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(d.) = 0
POL(f.0-0-0(x₁, x₂, x₃)) = 1
POL(f.0-0-1(x₁, x₂, x₃)) = 1
POL(f.0-1-0(x₁, x₂, x₃)) = 1
POL(f.0-1-1(x₁, x₂, x₃)) = x₂ + x₃
POL(f.1-0-0(x₁, x₂, x₃)) = 1
POL(f.1-0-1(x₁, x₂, x₃)) = 1 + x₁
POL(f.1-1-0(x₁, x₂, x₃)) = 1
POL(f.1-1-1(x₁, x₂, x₃)) = 1
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = 1
POL(mark.1(x₁)) = x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = 0
POL(proper.1(x₁)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
proper.1(b.) → ok.1(b.)
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))

(44) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(d.) = 0
POL(f.0-0-0(x₁, x₂, x₃)) = 1
POL(f.0-0-1(x₁, x₂, x₃)) = 0
POL(f.0-1-0(x₁, x₂, x₃)) = 1
POL(f.0-1-1(x₁, x₂, x₃)) = 0
POL(f.1-0-0(x₁, x₂, x₃)) = 1
POL(f.1-0-1(x₁, x₂, x₃)) = 0
POL(f.1-1-0(x₁, x₂, x₃)) = 1 + x₁
POL(f.1-1-1(x₁, x₂, x₃)) = 0
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = 1
POL(mark.1(x₁)) = x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = 0
POL(proper.1(x₁)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
proper.1(b.) → ok.1(b.)
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))

(46) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = x₁
POL(active.1(x₁)) = 0
POL(b.) = 1
POL(c.) = 0
POL(d.) = 1
POL(f.0-0-0(x₁, x₂, x₃)) = 1 + x₃
POL(f.0-0-1(x₁, x₂, x₃)) = 1
POL(f.0-1-0(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(f.0-1-1(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(f.1-0-0(x₁, x₂, x₃)) = 1 + x₁ + x₃
POL(f.1-0-1(x₁, x₂, x₃)) = 1 + x₁
POL(f.1-1-0(x₁, x₂, x₃)) = 1 + x₃
POL(f.1-1-1(x₁, x₂, x₃)) = 1
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = 1 + x₁
POL(mark.1(x₁)) = 1 + x₁
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = x₁
POL(proper.1(x₁)) = x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
proper.1(b.) → ok.1(b.)
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))

The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP.0(x₁)) = x₁
POL(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(d.) = 1
POL(f.0-0-0(x₁, x₂, x₃)) = x₁
POL(f.0-0-1(x₁, x₂, x₃)) = 0
POL(f.0-1-0(x₁, x₂, x₃)) = x₁
POL(f.0-1-1(x₁, x₂, x₃)) = 0
POL(f.1-0-0(x₁, x₂, x₃)) = x₁
POL(f.1-0-1(x₁, x₂, x₃)) = 0
POL(f.1-1-0(x₁, x₂, x₃)) = x₁
POL(f.1-1-1(x₁, x₂, x₃)) = 0
POL(m.1(x₁)) = 0
POL(mark.0(x₁)) = 0
POL(mark.1(x₁)) = x₁
POL(ok.0(x₁)) = 1 + x₁
POL(ok.1(x₁)) = 1 + x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))

(50) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

proper.1(b.) → ok.1(b.)
proper.0(c.) → ok.0(c.)
proper.0(d.) → ok.0(d.)
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.1-0-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
active.0(d.) → m.1(b.)
active.0(d.) → mark.0(c.)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(51) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) UsableRulesProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPSizeChangeProof (EQUIVALENT transformation)

(9) YES

(10) Obligation:

(11) UsableRulesProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPSizeChangeProof (EQUIVALENT transformation)

(14) YES

(15) Obligation:

(16) UsableRulesProof (EQUIVALENT transformation)

(17) Obligation:

(18) QDPSizeChangeProof (EQUIVALENT transformation)

(19) YES

(20) Obligation:

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

(22) Obligation:

(23) TransformationProof (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) Obligation:

(27) TransformationProof (EQUIVALENT transformation)

(28) Obligation:

(29) DependencyGraphProof (EQUIVALENT transformation)

(30) Obligation:

(31) SemLabProof (SOUND transformation)

(32) Obligation:

(33) DependencyGraphProof (EQUIVALENT transformation)

(34) Obligation:

(35) QDPOrderProof (EQUIVALENT transformation)

(36) Obligation:

(37) QDPOrderProof (EQUIVALENT transformation)

(38) Obligation:

(39) QDPOrderProof (EQUIVALENT transformation)

(40) Obligation:

(41) QDPOrderProof (EQUIVALENT transformation)

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) PisEmptyProof (EQUIVALENT transformation)

(52) YES