Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

TOP1(found1(x)) -> TOP1(active1(x))
CHECK1(f1(x)) -> CHECK1(x)
MATCH2(X, x) -> PROPER1(x)
ACTIVE1(f1(x)) -> ACTIVE1(x)
TOP1(mark1(x)) -> CHECK1(x)
ACTIVE1(f1(x)) -> F1(active1(x))
CHECK1(x) -> MATCH2(f1(X), x)
TOP1(mark1(x)) -> TOP1(check1(x))
CHECK1(f1(x)) -> F1(check1(x))
F1(mark1(x)) -> F1(x)
CHECK1(x) -> F1(X)
MATCH2(f1(x), f1(y)) -> F1(match2(x, y))
PROPER1(f1(x)) -> PROPER1(x)
TOP1(found1(x)) -> ACTIVE1(x)
F1(ok1(x)) -> F1(x)
CHECK1(x) -> START1(match2(f1(X), x))
TOP1(active1(c)) -> TOP1(mark1(c))
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
F1(found1(x)) -> F1(x)
PROPER1(f1(x)) -> F1(proper1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

TOP1(found1(x)) -> TOP1(active1(x))
CHECK1(f1(x)) -> CHECK1(x)
MATCH2(X, x) -> PROPER1(x)
ACTIVE1(f1(x)) -> ACTIVE1(x)
TOP1(mark1(x)) -> CHECK1(x)
ACTIVE1(f1(x)) -> F1(active1(x))
CHECK1(x) -> MATCH2(f1(X), x)
TOP1(mark1(x)) -> TOP1(check1(x))
CHECK1(f1(x)) -> F1(check1(x))
F1(mark1(x)) -> F1(x)
CHECK1(x) -> F1(X)
MATCH2(f1(x), f1(y)) -> F1(match2(x, y))
PROPER1(f1(x)) -> PROPER1(x)
TOP1(found1(x)) -> ACTIVE1(x)
F1(ok1(x)) -> F1(x)
CHECK1(x) -> START1(match2(f1(X), x))
TOP1(active1(c)) -> TOP1(mark1(c))
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
F1(found1(x)) -> F1(x)
PROPER1(f1(x)) -> F1(proper1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F1(mark1(x)) -> F1(x)
F1(ok1(x)) -> F1(x)
F1(found1(x)) -> F1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F1(mark1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.

F1(ok1(x)) -> F1(x)
F1(found1(x)) -> F1(x)
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = x1   
POL(found1(x1)) = x1   
POL(mark1(x1)) = 1 + 2·x1   
POL(ok1(x1)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F1(ok1(x)) -> F1(x)
F1(found1(x)) -> F1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F1(ok1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.

F1(found1(x)) -> F1(x)
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = x1   
POL(found1(x1)) = x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F1(found1(x)) -> F1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F1(found1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = 2·x1   
POL(found1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ACTIVE1(f1(x)) -> ACTIVE1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(f1(x)) -> ACTIVE1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(f1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

PROPER1(f1(x)) -> PROPER1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(f1(x)) -> PROPER1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(f1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

MATCH2(f1(x), f1(y)) -> MATCH2(x, y)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MATCH2(x1, x2)) = 2·x1 + 2·x2   
POL(f1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

CHECK1(f1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


CHECK1(f1(x)) -> CHECK1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CHECK1(x1)) = 2·x1   
POL(f1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

TOP1(found1(x)) -> TOP1(active1(x))
TOP1(active1(c)) -> TOP1(mark1(c))
TOP1(mark1(x)) -> TOP1(check1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(active1(c)) -> TOP1(mark1(c))
The remaining pairs can at least be oriented weakly.

TOP1(found1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(check1(x))
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = 2·x1   
POL(X) = 2   
POL(active1(x1)) = x1   
POL(c) = 2   
POL(check1(x1)) = 0   
POL(f1(x1)) = 0   
POL(found1(x1)) = x1   
POL(mark1(x1)) = 0   
POL(match2(x1, x2)) = 2·x1   
POL(ok1(x1)) = x1   
POL(proper1(x1)) = 2   
POL(start1(x1)) = 2·x1   

The following usable rules [14] were oriented:

active1(f1(x)) -> f1(active1(x))
f1(found1(x)) -> found1(f1(x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
f1(ok1(x)) -> ok1(f1(x))
check1(f1(x)) -> f1(check1(x))
start1(ok1(x)) -> found1(x)
f1(mark1(x)) -> mark1(f1(x))
check1(x) -> start1(match2(f1(X), x))
active1(f1(x)) -> mark1(x)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof

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

TOP1(found1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(check1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(mark1(x)) -> TOP1(check1(x))
The remaining pairs can at least be oriented weakly.

TOP1(found1(x)) -> TOP1(active1(x))
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = 2·x1   
POL(X) = 0   
POL(active1(x1)) = x1   
POL(c) = 0   
POL(check1(x1)) = x1   
POL(f1(x1)) = 2 + 2·x1   
POL(found1(x1)) = x1   
POL(mark1(x1)) = 2 + x1   
POL(match2(x1, x2)) = x2   
POL(ok1(x1)) = x1   
POL(proper1(x1)) = x1   
POL(start1(x1)) = x1   

The following usable rules [14] were oriented:

active1(f1(x)) -> f1(active1(x))
proper1(c) -> ok1(c)
f1(found1(x)) -> found1(f1(x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
f1(ok1(x)) -> ok1(f1(x))
check1(f1(x)) -> f1(check1(x))
match2(X, x) -> proper1(x)
start1(ok1(x)) -> found1(x)
f1(mark1(x)) -> mark1(f1(x))
proper1(f1(x)) -> f1(proper1(x))
check1(x) -> start1(match2(f1(X), x))
active1(f1(x)) -> mark1(x)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof

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

TOP1(found1(x)) -> TOP1(active1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(found1(x)) -> TOP1(active1(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = x1   
POL(active1(x1)) = 0   
POL(f1(x1)) = 2·x1   
POL(found1(x1)) = 2   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 2   

The following usable rules [14] were oriented:

active1(f1(x)) -> f1(active1(x))
f1(found1(x)) -> found1(f1(x))
f1(ok1(x)) -> ok1(f1(x))
f1(mark1(x)) -> mark1(f1(x))
active1(f1(x)) -> mark1(x)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.