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, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> x

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> 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(s, x)) -> A2(p, a2(s, x))
A2(f, a2(s, x)) -> A2(d, a2(f, a2(p, a2(s, x))))
A2(d, a2(s, x)) -> A2(d, a2(p, a2(s, x)))
A2(d, a2(s, x)) -> A2(s, a2(d, a2(p, a2(s, x))))
A2(d, a2(s, x)) -> A2(s, a2(s, a2(d, a2(p, a2(s, x)))))
A2(f, a2(s, x)) -> A2(f, a2(p, a2(s, x)))
A2(f, 0) -> A2(s, 0)
A2(d, a2(s, x)) -> A2(p, a2(s, x))

The TRS R consists of the following rules:

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> 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(s, x)) -> A2(p, a2(s, x))
A2(f, a2(s, x)) -> A2(d, a2(f, a2(p, a2(s, x))))
A2(d, a2(s, x)) -> A2(d, a2(p, a2(s, x)))
A2(d, a2(s, x)) -> A2(s, a2(d, a2(p, a2(s, x))))
A2(d, a2(s, x)) -> A2(s, a2(s, a2(d, a2(p, a2(s, x)))))
A2(f, a2(s, x)) -> A2(f, a2(p, a2(s, x)))
A2(f, 0) -> A2(s, 0)
A2(d, a2(s, x)) -> A2(p, a2(s, x))

The TRS R consists of the following rules:

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 6 less nodes.

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

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

A2(d, a2(s, x)) -> A2(d, a2(p, a2(s, x)))

The TRS R consists of the following rules:

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> 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(d, a2(s, x)) -> A2(d, a2(p, a2(s, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( d ) = 1


POL( s ) = 3


POL( A2(x1, x2) ) = max{0, x2 - 3}


POL( p ) = max{0, -2}


POL( a2(x1, x2) ) = max{0, 2x1 + x2 - 2}



The following usable rules [14] were oriented:

a2(p, a2(s, x)) -> x



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

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

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> 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.

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

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

A2(f, a2(s, x)) -> A2(f, a2(p, a2(s, x)))

The TRS R consists of the following rules:

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> 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(s, x)) -> A2(f, a2(p, a2(s, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( f ) = 1


POL( s ) = 3


POL( A2(x1, x2) ) = max{0, x2 - 3}


POL( p ) = max{0, -2}


POL( a2(x1, x2) ) = max{0, 2x1 + x2 - 2}



The following usable rules [14] were oriented:

a2(p, a2(s, x)) -> x



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof

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

a2(f, 0) -> a2(s, 0)
a2(d, 0) -> 0
a2(d, a2(s, x)) -> a2(s, a2(s, a2(d, a2(p, a2(s, x)))))
a2(f, a2(s, x)) -> a2(d, a2(f, a2(p, a2(s, x))))
a2(p, a2(s, x)) -> 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.