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

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

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

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

A(h, x) → A(f, x)
A(f, a(f, x)) → A(x, 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 [15].


The following pairs can be oriented strictly and are deleted.


A(f, a(f, x)) → A(x, x)
The remaining pairs can at least be oriented weakly.

A(h, x) → A(f, x)
Used ordering: Polynomial interpretation [25,35]:

POL(f) = 0   
POL(a(x1, x2)) = 1/4 + (4)x_2   
POL(A(x1, x2)) = (4)x_2   
POL(h) = 0   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

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

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.