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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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)
G2(s1(x), y) -> +12(y, s1(x))
G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
F1(s1(x)) -> G2(x, s1(x))
G2(s1(x), y) -> +12(y, x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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)
G2(s1(x), y) -> +12(y, s1(x))
G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
F1(s1(x)) -> G2(x, s1(x))
G2(s1(x), y) -> +12(y, x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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 3 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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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 interpretation [21]:

POL(+12(x1, x2)) = x2   
POL(s1(x1)) = 1 + x1   

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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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:

G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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.


G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + x1 + x2   
POL(0) = 0   
POL(G2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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.