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.

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 contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

A2(f, a2(f, x)) -> A2(x, x)
Used argument filtering: A2(x1, x2)  =  x2
a2(x1, x2)  =  a1(x2)
f  =  f
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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 contains 0 SCCs with 1 less node.