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:

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))

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:

G(g(x)) → G(h(g(x)))
H(h(x)) → H(f(h(x), x))
G(g(x)) → H(g(x))

The TRS R consists of the following rules:

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), 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:

G(g(x)) → G(h(g(x)))
H(h(x)) → H(f(h(x), x))
G(g(x)) → H(g(x))

The TRS R consists of the following rules:

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

G(g(x)) → G(h(g(x)))

The TRS R consists of the following rules:

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))

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.


G(g(x)) → G(h(g(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( g(x1) ) =
/0\
\1/
+
/11\
\11/
·x1

M( f(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( h(x1) ) =
/0\
\0/
+
/11\
\00/
·x1

Tuple symbols:
M( G(x1) ) = 0+
[0,1]
·x1


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:

g(g(x)) → g(h(g(x)))
g(h(g(x))) → g(x)
h(h(x)) → h(f(h(x), x))



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

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

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), 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.