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:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
F(x, y, g(z)) → F(x, y, z)

The TRS R consists of the following rules:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
F(x, y, g(z)) → F(x, y, z)

The TRS R consists of the following rules:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, y, g(z)) → F(x, y, z)
The remaining pairs can at least be oriented weakly.

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( g(x1) ) =
/1\
\0/
+
/10\
\00/
·x1

M( 0 ) =
/0\
\0/

M( 1 ) =
/0\
\0/

Tuple symbols:
M( F(x1, ..., x3) ) = 0+
[0,0]
·x1+
[0,0]
·x2+
[1,0]
·x3


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof

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

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)

The TRS R consists of the following rules:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.f: 0
g: x0
0: 1
1: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)
F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)
F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.1-0-1(0., 1., x) → F.1-1-1(g.1(x), g.1(x), x)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)
F.1-0-0(0., 1., x) → F.0-0-0(g.0(x), g.0(x), x)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof

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

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)
F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)
F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.1-0-1(0., 1., x) → F.1-1-1(g.1(x), g.1(x), x)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)
F.1-0-0(0., 1., x) → F.0-0-0(g.0(x), g.0(x), x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 8 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-1-1(x1, x2, x3)) = x1 + x2 + x3   
POL(g.1(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-1-0(x1, x2, x3)) = x1 + x2 + x3   
POL(g.1(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-0-1(x1, x2, x3)) = x1 + x2 + x3   
POL(g.0(x1)) = x1   
POL(g.1(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-1-1(x1, x2, x3)) = x1 + x2 + x3   
POL(g.0(x1)) = x1   
POL(g.1(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-1-0(x1, x2, x3)) = x1 + x2 + x3   
POL(g.0(x1)) = x1   
POL(g.1(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP

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

F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0-1(x1, x2, x3)) = x1 + x2 + x3   
POL(g.0(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP

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

F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0-0(x1, x2, x3)) = x1 + x2 + x3   
POL(g.0(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof

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

F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.

The following dependency pairs can be deleted:

F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-0-0(x1, x2, x3)) = x1 + x2 + x3   
POL(g.0(x1)) = x1   
POL(g.1(x1)) = x1   



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

Q DP problem:
P is empty.
R is empty.
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.