Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
SUM(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIBO(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
LT(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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