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(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(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:

F(f(x)) → F(c(f(x)))
F(f(x)) → F(d(f(x)))
G(c(h(0))) → G(d(1))
G(h(x)) → G(x)
G(c(1)) → G(d(h(0)))

The TRS R consists of the following rules:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(f(x)) → F(c(f(x)))
F(f(x)) → F(d(f(x)))
G(c(h(0))) → G(d(1))
G(h(x)) → G(x)
G(c(1)) → G(d(h(0)))

The TRS R consists of the following rules:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(f(x)) → F(c(f(x)))
F(f(x)) → F(d(f(x)))
G(c(h(0))) → G(d(1))
G(h(x)) → G(x)
G(c(1)) → G(d(h(0)))

The TRS R consists of the following rules:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

G(h(x)) → G(x)

The TRS R consists of the following rules:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(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.


G(h(x)) → G(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
h(x1)  =  h(x1)

Recursive path order with status [2].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented: none



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

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

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(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.