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:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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:

LT(s(x), s(y)) → LT(x, y)
IF(true, b, c, x, y) → PLUS(x, y)
FIBITER(b, c, x, y) → LT(c, b)
FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y)
IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
FIB(x) → FIBITER(x, 0, 0, s(0))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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:

LT(s(x), s(y)) → LT(x, y)
IF(true, b, c, x, y) → PLUS(x, y)
FIBITER(b, c, x, y) → LT(c, b)
FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y)
IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
FIB(x) → FIBITER(x, 0, 0, s(0))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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 3 SCCs with 3 less nodes.

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

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x1, x2)) = (2)x_1   
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
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(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

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

FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y)
IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(x, y))

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