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:

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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:

G2(h, s1(s1(x))) -> G2(h, x)
G2(d, s1(x)) -> G2(d, x)
ID1(x) -> F2(x, s1(0))
F2(s1(x), y) -> DOUBLE1(y)
F2(s1(x), y) -> F2(half1(s1(x)), double1(y))
HALF1(x) -> G2(h, x)
F2(s1(x), y) -> HALF1(s1(x))
DOUBLE1(x) -> G2(d, x)

The TRS R consists of the following rules:

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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:

G2(h, s1(s1(x))) -> G2(h, x)
G2(d, s1(x)) -> G2(d, x)
ID1(x) -> F2(x, s1(0))
F2(s1(x), y) -> DOUBLE1(y)
F2(s1(x), y) -> F2(half1(s1(x)), double1(y))
HALF1(x) -> G2(h, x)
F2(s1(x), y) -> HALF1(s1(x))
DOUBLE1(x) -> G2(d, x)

The TRS R consists of the following rules:

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 5 less nodes.

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

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

G2(h, s1(s1(x))) -> G2(h, x)

The TRS R consists of the following rules:

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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.


G2(h, s1(s1(x))) -> G2(h, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( G2(x1, x2) ) = max{0, x2 - 1}


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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

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

G2(d, s1(x)) -> G2(d, x)

The TRS R consists of the following rules:

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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.


G2(d, s1(x)) -> G2(d, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( G2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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

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

F2(s1(x), y) -> F2(half1(s1(x)), double1(y))

The TRS R consists of the following rules:

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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.


F2(s1(x), y) -> F2(half1(s1(x)), double1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F2(x1, x2) ) = x1


POL( s1(x1) ) = x1 + 1


POL( half1(x1) ) = max{0, x1 - 1}


POL( g2(x1, x2) ) = max{0, x2 - 1}


POL( 0 ) = 0



The following usable rules [14] were oriented:

g2(x, 0) -> 0
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
half1(x) -> g2(h, x)



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

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

g2(x, 0) -> 0
g2(d, s1(x)) -> s1(s1(g2(d, x)))
g2(h, s1(0)) -> 0
g2(h, s1(s1(x))) -> s1(g2(h, x))
double1(x) -> g2(d, x)
half1(x) -> g2(h, x)
f2(s1(x), y) -> f2(half1(s1(x)), double1(y))
f2(s1(0), y) -> y
id1(x) -> f2(x, s1(0))

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.