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:

a2(f, a2(f, x)) -> a2(x, x)
a2(h, x) -> a2(f, a2(g, a2(f, x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a2(f, a2(f, x)) -> a2(x, x)
a2(h, x) -> a2(f, a2(g, a2(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:

A2(f, a2(f, x)) -> A2(x, x)
A2(h, x) -> A2(f, x)
A2(h, x) -> A2(f, a2(g, a2(f, x)))
A2(h, x) -> A2(g, a2(f, x))

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, x)
a2(h, x) -> a2(f, a2(g, a2(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:

A2(f, a2(f, x)) -> A2(x, x)
A2(h, x) -> A2(f, x)
A2(h, x) -> A2(f, a2(g, a2(f, x)))
A2(h, x) -> A2(g, a2(f, x))

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, x)
a2(h, x) -> a2(f, a2(g, a2(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

A2(f, a2(f, x)) -> A2(x, x)
A2(h, x) -> A2(f, x)

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, x)
a2(h, x) -> a2(f, a2(g, a2(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.


A2(f, a2(f, x)) -> A2(x, x)
The remaining pairs can at least be oriented weakly.

A2(h, x) -> A2(f, x)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( f ) = 1


POL( A2(x1, x2) ) = x2


POL( a2(x1, x2) ) = x1 + x2 + 1


POL( h ) = 1



The following usable rules [14] were oriented: none



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

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

A2(h, x) -> A2(f, x)

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, x)
a2(h, x) -> a2(f, a2(g, a2(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 0 SCCs with 1 less node.