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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, 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 2 less nodes.

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

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

A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, x)

The TRS R consists of the following rules:

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


A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1, x2)  =  A(x1, x2)
f  =  f
a(x1, x2)  =  a(x2)
h  =  h

Lexicographic path order with status [19].
Precedence:
a1 > A2
h > f > A2

Status:
A2: [1,2]
a1: multiset
f: multiset
h: multiset

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:

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