Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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:

F(s(x1)) → G(s(x1))
G(x1) → HALF(x1)
HALF(s(s(x1))) → P(p(s(s(x1))))
HALF(s(s(x1))) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
HALF(0(x1)) → 01(s(s(half(x1))))
RD(0(x1)) → RD(x1)
RD(0(x1)) → 01(0(0(rd(x1))))
P(0(x1)) → P(x1)
HALF(0(x1)) → HALF(x1)
RD(0(x1)) → 01(0(0(0(0(rd(x1))))))
P(0(x1)) → 01(s(s(p(x1))))
RD(0(x1)) → 01(0(rd(x1)))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
G(x1) → I(s(half(x1)))
I(x1) → P(x1)
RD(0(x1)) → 01(rd(x1))
RD(0(x1)) → 01(0(0(0(0(0(rd(x1)))))))
I(x1) → F(p(x1))
RD(0(x1)) → 01(0(0(0(rd(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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:

F(s(x1)) → G(s(x1))
G(x1) → HALF(x1)
HALF(s(s(x1))) → P(p(s(s(x1))))
HALF(s(s(x1))) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
HALF(0(x1)) → 01(s(s(half(x1))))
RD(0(x1)) → RD(x1)
RD(0(x1)) → 01(0(0(rd(x1))))
P(0(x1)) → P(x1)
HALF(0(x1)) → HALF(x1)
RD(0(x1)) → 01(0(0(0(0(rd(x1))))))
P(0(x1)) → 01(s(s(p(x1))))
RD(0(x1)) → 01(0(rd(x1)))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
G(x1) → I(s(half(x1)))
I(x1) → P(x1)
RD(0(x1)) → 01(rd(x1))
RD(0(x1)) → 01(0(0(0(0(0(rd(x1)))))))
I(x1) → F(p(x1))
RD(0(x1)) → 01(0(0(0(rd(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 12 less nodes.

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

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

RD(0(x1)) → RD(x1)

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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.


RD(0(x1)) → RD(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(RD(x1)) = (4)x_1   
POL(0(x1)) = 1 + (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



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

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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
          ↳ QDP
          ↳ QDP

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

P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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.


P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (4)x_1   
POL(p(x1)) = 4 + (4)x_1   
POL(s(x1)) = 4 + (4)x_1   
POL(0(x1)) = 4 + (4)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



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

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

HALF(0(x1)) → HALF(x1)
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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.


HALF(0(x1)) → HALF(x1)
The remaining pairs can at least be oriented weakly.

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x1)) = x_1   
POL(s(x1)) = x_1   
POL(p(x1)) = x_1   
POL(0(x1)) = 2 + (4)x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
0(x1) → x1



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

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

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

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

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.