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(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(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(a(x)) → F(a(x), b)
A(a(x)) → F(b, a(f(a(x), b)))
A(a(f(b, a(x)))) → F(a(a(a(x))), b)
A(a(f(b, a(x)))) → A(a(x))
A(a(x)) → A(f(a(x), b))
A(a(f(b, a(x)))) → A(a(a(x)))
F(a(x), b) → F(b, a(x))

The TRS R consists of the following rules:

a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(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(a(x)) → F(a(x), b)
A(a(x)) → F(b, a(f(a(x), b)))
A(a(f(b, a(x)))) → F(a(a(a(x))), b)
A(a(f(b, a(x)))) → A(a(x))
A(a(x)) → A(f(a(x), b))
A(a(f(b, a(x)))) → A(a(a(x)))
F(a(x), b) → F(b, a(x))

The TRS R consists of the following rules:

a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(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 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A(a(f(b, a(x)))) → A(a(x))


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b) = 0   
POL(f(x1, x2)) = x1 + x2   



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

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

A(a(f(b, a(x)))) → A(a(a(x)))

The TRS R consists of the following rules:

a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(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(a(f(b, a(x)))) → A(a(a(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( f(x1, x2) ) =
/0\
\1/
+
/00\
\01/
·x1+
/00\
\01/
·x2

M( a(x1) ) =
/1\
\0/
+
/01\
\10/
·x1

M( b ) =
/0\
\0/

Tuple symbols:
M( A(x1) ) = 0+
[1,0]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

f(a(x), b) → f(b, a(x))
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(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.