Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
      ↳26 Instantiation (⇔)
        ↳27 QDP
      ↳28 NonInfProof (⇔)
        ↳29 QDP
        ↳30 DependencyGraphProof (⇔)
        ↳31 TRUE

(0) Obligation:

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.

(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:

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))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

(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:

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

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) 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:

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))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

(11) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LT(s(x), s(y)) → LT(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

(18) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • PLUS(s(x), y) → PLUS(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) Obligation:

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

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

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

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

The TRS R consists of the following rules:

lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fib(x0)
fibiter(x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
if(true, x0, x1, x2, x3)

(25) Obligation:

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

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

The TRS R consists of the following rules:

lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y) we obtained the following new rules [LPAR04]:

FIBITER(z0, s(z1), z3, y_0) → IF(lt(s(z1), z0), z0, s(z1), z3, y_0)

(27) Obligation:

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

IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
FIBITER(z0, s(z1), z3, y_0) → IF(lt(s(z1), z0), z0, s(z1), z3, y_0)

The TRS R consists of the following rules:

lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(28) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y)) the following chains were created:
  • We consider the chain 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)) which results in the following constraint:

    (1)    (IF(lt(x5, x4), x4, x5, x6, x7)=IF(true, x8, x9, x10, x11) ⇒ IF(true, x8, x9, x10, x11)≥FIBITER(x8, s(x9), x11, plus(x10, x11)))



    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

    (2)    (lt(x5, x4)=true ⇒ IF(true, x4, x5, x6, x7)≥FIBITER(x4, s(x5), x7, plus(x6, x7)))



    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x5, x4)=true which results in the following new constraints:

    (3)    (true=true ⇒ IF(true, s(x24), 0, x6, x7)≥FIBITER(s(x24), s(0), x7, plus(x6, x7)))


    (4)    (lt(x27, x26)=true∧(∀x28,x29:lt(x27, x26)=true ⇒ IF(true, x26, x27, x28, x29)≥FIBITER(x26, s(x27), x29, plus(x28, x29))) ⇒ IF(true, s(x26), s(x27), x6, x7)≥FIBITER(s(x26), s(s(x27)), x7, plus(x6, x7)))



    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

    (5)    (IF(true, s(x24), 0, x6, x7)≥FIBITER(s(x24), s(0), x7, plus(x6, x7)))



    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x28,x29:lt(x27, x26)=true ⇒ IF(true, x26, x27, x28, x29)≥FIBITER(x26, s(x27), x29, plus(x28, x29))) with σ = [x28 / x6, x29 / x7] which results in the following new constraint:

    (6)    (IF(true, x26, x27, x6, x7)≥FIBITER(x26, s(x27), x7, plus(x6, x7)) ⇒ IF(true, s(x26), s(x27), x6, x7)≥FIBITER(s(x26), s(s(x27)), x7, plus(x6, x7)))







For Pair FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y) the following chains were created:
  • We consider the chain IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y)), FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y) which results in the following constraint:

    (7)    (FIBITER(x12, s(x13), x15, plus(x14, x15))=FIBITER(x16, x17, x18, x19) ⇒ FIBITER(x16, x17, x18, x19)≥IF(lt(x17, x16), x16, x17, x18, x19))



    We simplified constraint (7) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (8)    (FIBITER(x12, s(x13), x15, x19)≥IF(lt(s(x13), x12), x12, s(x13), x15, x19))







To summarize, we get the following constraints P≥ for the following pairs.
  • IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
    • (IF(true, s(x24), 0, x6, x7)≥FIBITER(s(x24), s(0), x7, plus(x6, x7)))
    • (IF(true, x26, x27, x6, x7)≥FIBITER(x26, s(x27), x7, plus(x6, x7)) ⇒ IF(true, s(x26), s(x27), x6, x7)≥FIBITER(s(x26), s(s(x27)), x7, plus(x6, x7)))

  • FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y)
    • (FIBITER(x12, s(x13), x15, x19)≥IF(lt(s(x13), x12), x12, s(x13), x15, x19))




The constraints for P> respective Pbound are constructed from P≥ where we just replace every occurence of "t ≥ s" in P≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(FIBITER(x1, x2, x3, x4)) = -1 + x1 - x2   
POL(IF(x1, x2, x3, x4, x5)) = -1 + x2 - x3   
POL(c) = -2   
POL(false) = 0   
POL(lt(x1, x2)) = 0   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
The following pairs are in Pbound:

IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
There are no usable rules

(29) Obligation:

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

FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y)

The TRS R consists of the following rules:

lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

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

(30) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(31) TRUE