Termination w.r.t. Q proof of /tmp/tmpRiYxkz/otto04.trs

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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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:

FIBO(s(s(x))) → FIBO(x)
FIBO(s(0)) → FIB(s(0))
IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
FIBO(0) → FIB(0)
FIBO(s(s(x))) → FIBO(s(x))
IF(false, c, s(s(x)), a, b) → SUM(fibo(a), fibo(b))
LT(s(x), s(y)) → LT(x, y)
FIB(s(s(x))) → IF(true, 0, s(s(x)), 0, 0)
FIBO(s(s(x))) → SUM(fibo(s(x)), fibo(x))
IF(false, c, s(s(x)), a, b) → FIBO(b)
SUM(x, s(y)) → SUM(x, y)
IF(false, c, s(s(x)), a, b) → FIBO(a)
IF(true, c, s(s(x)), a, b) → LT(s(c), s(s(x)))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(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:

FIBO(s(s(x))) → FIBO(x)
FIBO(s(0)) → FIB(s(0))
IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
FIBO(0) → FIB(0)
FIBO(s(s(x))) → FIBO(s(x))
IF(false, c, s(s(x)), a, b) → SUM(fibo(a), fibo(b))
LT(s(x), s(y)) → LT(x, y)
FIB(s(s(x))) → IF(true, 0, s(s(x)), 0, 0)
FIBO(s(s(x))) → SUM(fibo(s(x)), fibo(x))
IF(false, c, s(s(x)), a, b) → FIBO(b)
SUM(x, s(y)) → SUM(x, y)
IF(false, c, s(s(x)), a, b) → FIBO(a)
IF(true, c, s(s(x)), a, b) → LT(s(c), s(s(x)))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SUM(x, s(y)) → SUM(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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.

SUM(x, s(y)) → SUM(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(SUM(x₁, x₂)) = (2)x_2
POL(s(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(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

          ↳ QDP

          ↳ QDP

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

FIBO(s(s(x))) → FIBO(x)
FIBO(s(s(x))) → FIBO(s(x))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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.

FIBO(s(s(x))) → FIBO(x)
FIBO(s(s(x))) → FIBO(s(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(FIBO(x₁)) = (4)x_1
POL(s(x₁)) = 1/4 + (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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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.

LT(s(x), s(y)) → LT(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 1/2 + (13/4)x_1
POL(LT(x₁, x₂)) = (15/4)x_2

The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(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

          ↳ QDP

          ↳ QDP

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

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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