Termination w.r.t. Q proof of /tmp/tmphSBppt/Ex24_Luc06

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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, X, c)) → F(X, c, X)
ACTIVE(f(X1, X2, X3)) → F(X1, active(X2), X3)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)
F(X1, mark(X2), X3) → F(X1, X2, X3)
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)
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)

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x2)
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = x1
f(x1, x2, x3) = f(x2, x3)
b = b
c = c
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[c, proper₁] > b > f₂ > [F₁, ok₁]

The following usable rules [FROCOS05] were oriented:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

(7) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(X1, mark(X2), X3) → F(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = x2
mark(x1) = mark(x1)
active(x1) = active
f(x1, x2, x3) = x2
b = b
c = c
proper(x1) = proper
ok(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active > [c, proper, top] > [mark₁, b]

The following usable rules [FROCOS05] were oriented:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

(9) Obligation:

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

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
f(x1, x2, x3) = f(x1, x2, x3)
active(x1) = x1
b = b
c = c
mark(x1) = mark
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Recursive Path Order [RPO].
Precedence:

b > [c, mark, ok] > f₃ > PROPER₁

The following usable rules [FROCOS05] were oriented:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

(14) Obligation:

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

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
f(x1, x2, x3) = f(x2)
active(x1) = x1
b = b
c = c
mark(x1) = x1
proper(x1) = x1
ok(x1) = x1
top(x1) = x1

Recursive Path Order [RPO].
Precedence:

f₁ > ACTIVE₁ > [b, c]

The following usable rules [FROCOS05] were oriented:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

(19) Obligation:

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

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(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:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
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))

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