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

(0) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

TOP(active(c)) → TOP(mark(c))
TOP(mark(x)) → TOP(check(x))
TOP(mark(x)) → CHECK(x)
CHECK(f(x)) → F(check(x))
CHECK(f(x)) → CHECK(x)
CHECK(x) → START(match(f(X), x))
CHECK(x) → MATCH(f(X), x)
CHECK(x) → F(X)
MATCH(f(x), f(y)) → F(match(x, y))
MATCH(f(x), f(y)) → MATCH(x, y)
MATCH(X, x) → PROPER(x)
PROPER(f(x)) → F(proper(x))
PROPER(f(x)) → PROPER(x)
F(ok(x)) → F(x)
F(found(x)) → F(x)
TOP(found(x)) → TOP(active(x))
TOP(found(x)) → ACTIVE(x)
ACTIVE(f(x)) → F(active(x))
ACTIVE(f(x)) → ACTIVE(x)
F(mark(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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 6 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(found(x)) → F(x)
F(ok(x)) → F(x)
F(mark(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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(found(x)) → F(x)
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)  =  F(x1)
found(x1)  =  found(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1)  =  x1
top(x1)  =  top(x1)
c  =  c
check(x1)  =  check(x1)
start(x1)  =  start(x1)
match(x1, x2)  =  match
X  =  X
proper(x1)  =  proper

Lexicographic Path Order [LPO].
Precedence:

top₁ > [F₁, found₁, mark₁, active₁, check₁, start₁] > X > [c, match, proper]

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(7) Obligation:

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

F(ok(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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(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)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
mark(x1)  =  mark
top(x1)  =  top
c  =  c
check(x1)  =  check(x1)
start(x1)  =  start
match(x1, x2)  =  match(x2)
X  =  X
proper(x1)  =  proper(x1)
found(x1)  =  found

Lexicographic Path Order [LPO].
Precedence:

[active₁, top, c, match₁, proper₁] > [F₁, ok₁, f₁, mark, check₁] > [start, found]

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(9) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

ACTIVE(f(x)) → ACTIVE(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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.

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)  =  ACTIVE(x1)
f(x1)  =  f(x1)
active(x1)  =  active(x1)
mark(x1)  =  mark
top(x1)  =  top
c  =  c
check(x1)  =  x1
start(x1)  =  x1
match(x1, x2)  =  x2
X  =  X
proper(x1)  =  x1
ok(x1)  =  x1
found(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

[f₁, active₁] > [ACTIVE₁, mark, top, c]
X > [ACTIVE₁, mark, top, c]

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(14) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

PROPER(f(x)) → PROPER(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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.

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)  =  PROPER(x1)
f(x1)  =  f(x1)
active(x1)  =  active(x1)
mark(x1)  =  mark
top(x1)  =  top
c  =  c
check(x1)  =  x1
start(x1)  =  x1
match(x1, x2)  =  x2
X  =  X
proper(x1)  =  x1
ok(x1)  =  x1
found(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

[f₁, active₁] > [PROPER₁, mark, top, c]
X > [PROPER₁, mark, top, c]

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(19) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

MATCH(f(x), f(y)) → MATCH(x, y)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MATCH(f(x), f(y)) → MATCH(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MATCH(x1, x2)  =  x2
f(x1)  =  f(x1)
active(x1)  =  x1
mark(x1)  =  x1
top(x1)  =  top
c  =  c
check(x1)  =  check(x1)
start(x1)  =  x1
match(x1, x2)  =  match(x2)
X  =  X
proper(x1)  =  proper(x1)
ok(x1)  =  x1
found(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

[top, check₁] > c
[top, check₁] > match₁ > proper₁ > f₁
[top, check₁] > X

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(24) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

CHECK(f(x)) → CHECK(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CHECK(f(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  CHECK(x1)
f(x1)  =  f(x1)
active(x1)  =  active(x1)
mark(x1)  =  mark
top(x1)  =  top
c  =  c
check(x1)  =  x1
start(x1)  =  x1
match(x1, x2)  =  x2
X  =  X
proper(x1)  =  x1
ok(x1)  =  x1
found(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

[f₁, active₁] > [CHECK₁, mark, top, c]
X > [CHECK₁, mark, top, c]

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(29) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
TOP(active(c)) → TOP(mark(c))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(active(c)) → TOP(mark(c))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
mark(x1)  =  mark
check(x1)  =  check
found(x1)  =  x1
active(x1)  =  x1
c  =  c
f(x1)  =  f
top(x1)  =  top
start(x1)  =  x1
match(x1, x2)  =  x1
X  =  X
proper(x1)  =  proper
ok(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

TOP₁ > [c, proper] > top > [mark, check, f]
X > [c, proper] > top > [mark, check, f]

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(34) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(found(x)) → TOP(active(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
mark(x1)  =  mark(x1)
check(x1)  =  check(x1)
found(x1)  =  found(x1)
active(x1)  =  x1
f(x1)  =  f(x1)
top(x1)  =  top
c  =  c
start(x1)  =  start(x1)
match(x1, x2)  =  x2
X  =  X
proper(x1)  =  x1
ok(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

[mark₁, check₁, found₁, f₁, top, c, start₁] > TOP₁
[mark₁, check₁, found₁, f₁, top, c, start₁] > X

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(36) Obligation:

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

TOP(mark(x)) → TOP(check(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(mark(x)) → TOP(check(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
mark(x1)  =  mark(x1)
check(x1)  =  x1
active(x1)  =  x1
f(x1)  =  f(x1)
top(x1)  =  top
c  =  c
start(x1)  =  x1
match(x1, x2)  =  x2
X  =  X
proper(x1)  =  x1
ok(x1)  =  x1
found(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

f₁ > [mark₁, top] > TOP₁
f₁ > [mark₁, top] > c

The following usable rules [FROCOS05] were oriented:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

(38) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE