(0) Obligation:

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(X)) → IF(X, c, f(true))
ACTIVE(f(X)) → F(true)
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) → IF(X1, active(X2), X3)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X2)
F(mark(X)) → F(X)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
PROPER(f(X)) → F(proper(X))
PROPER(f(X)) → PROPER(X)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
F(ok(X)) → F(X)
IF(ok(X1), ok(X2), ok(X3)) → IF(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(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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 5 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.


IF(X1, mark(X2), X3) → IF(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  IF(x2)
mark(x1)  =  mark(x1)
ok(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
IF1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.


IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  x2
mark(x1)  =  x1
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF(mark(X1), X2, X3) → IF(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  x1
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

F(ok(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(ok(X)) → F(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
ok1 > F1

Status:
F1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(mark(X)) → F(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(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)
if(x1, x2, x3)  =  if(x1, x2, x3)
f(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
PROPER1: [1]
if3: [1,2,3]

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(f(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
f(x1)  =  f(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
f1: [1]

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(if(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)
if(x1, x2, x3)  =  if(x1, x2)
f(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ACTIVE1: [1]
if2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(f(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
f(x1)  =  f(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
f1: [1]

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) 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(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.