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:

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(x)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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:

+12(x, s1(y)) -> +12(x, y)
H1(x) -> Q1(x)
H1(x) -> P1(x)
G1(s1(x)) -> G1(x)
F1(s1(s1(x))) -> +12(p1(g1(x)), q1(g1(x)))
F1(s1(s1(x))) -> G1(x)
G1(s1(x)) -> P1(g1(x))
F1(s1(s1(x))) -> H1(g1(x))
F1(s1(s1(x))) -> P1(g1(x))
G1(s1(x)) -> H1(g1(x))
G1(s1(x)) -> +12(p1(g1(x)), q1(g1(x)))
F1(s1(s1(x))) -> P1(h1(g1(x)))
F1(s1(s1(x))) -> Q1(g1(x))
H1(x) -> +12(p1(x), q1(x))
G1(s1(x)) -> Q1(g1(x))

The TRS R consists of the following rules:

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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:

+12(x, s1(y)) -> +12(x, y)
H1(x) -> Q1(x)
H1(x) -> P1(x)
G1(s1(x)) -> G1(x)
F1(s1(s1(x))) -> +12(p1(g1(x)), q1(g1(x)))
F1(s1(s1(x))) -> G1(x)
G1(s1(x)) -> P1(g1(x))
F1(s1(s1(x))) -> H1(g1(x))
F1(s1(s1(x))) -> P1(g1(x))
G1(s1(x)) -> H1(g1(x))
G1(s1(x)) -> +12(p1(g1(x)), q1(g1(x)))
F1(s1(s1(x))) -> P1(h1(g1(x)))
F1(s1(s1(x))) -> Q1(g1(x))
H1(x) -> +12(p1(x), q1(x))
G1(s1(x)) -> Q1(g1(x))

The TRS R consists of the following rules:

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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 13 less nodes.

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

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

+12(x, s1(y)) -> +12(x, y)

The TRS R consists of the following rules:

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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.


+12(x, s1(y)) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = 2x1 + 2x2 + 2


POL( s1(x1) ) = 3x1 + 1



The following usable rules [14] were oriented: none



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

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

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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:

G1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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.


G1(s1(x)) -> G1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 3x1 + 3


POL( G1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



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

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

f1(0) -> 0
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> p1(h1(g1(x)))
g1(0) -> pair2(s1(0), s1(0))
g1(s1(x)) -> h1(g1(x))
h1(x) -> pair2(+2(p1(x), q1(x)), p1(x))
p1(pair2(x, y)) -> x
q1(pair2(x, y)) -> y
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
f1(s1(s1(x))) -> +2(p1(g1(x)), q1(g1(x)))
g1(s1(x)) -> pair2(+2(p1(g1(x)), q1(g1(x))), p1(g1(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.