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:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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)
FIB1(s1(s1(x))) -> G1(x)
NP1(pair2(x, y)) -> +12(x, y)
SP1(pair2(x, y)) -> +12(x, y)
FIB1(s1(s1(x))) -> SP1(g1(x))
G1(s1(x)) -> G1(x)
G1(s1(x)) -> NP1(g1(x))

The TRS R consists of the following rules:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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)
FIB1(s1(s1(x))) -> G1(x)
NP1(pair2(x, y)) -> +12(x, y)
SP1(pair2(x, y)) -> +12(x, y)
FIB1(s1(s1(x))) -> SP1(g1(x))
G1(s1(x)) -> G1(x)
G1(s1(x)) -> NP1(g1(x))

The TRS R consists of the following rules:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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 5 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:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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)) = 3·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:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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 interpretation [21]:

POL(G1(x1)) = 3·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:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(0))) -> s1(0)
fib1(s1(s1(x))) -> sp1(g1(x))
g1(0) -> pair2(s1(0), 0)
g1(s1(0)) -> pair2(s1(0), s1(0))
g1(s1(x)) -> np1(g1(x))
sp1(pair2(x, y)) -> +2(x, y)
np1(pair2(x, y)) -> pair2(+2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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.