Termination w.r.t. Q proof of /tmp/tmpHrFAJy/thiemann25.trs

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:

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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))

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)

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))

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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))

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.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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))

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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))

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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))

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

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:

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.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.

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

We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint:
(2) (FIBITER(x4, s(x5), x7, x11)≥IF(lt(s(x5), x4), x4, s(x5), x7, x11))

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

We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(4)    (lt(x13, x12)=true ⇒ IF(true, x12, x13, x14, x15)≥FIBITER(x12, s(x13), x15, plus(x14, x15)))

We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x13, x12)=true which results in the following new constraints:
(5)    (true=true ⇒ IF(true, s(x24), 0, x14, x15)≥FIBITER(s(x24), s(0), x15, plus(x14, x15)))

(6)    (lt(x26, x27)=true∧(∀x28,x29:lt(x26, x27)=true ⇒ IF(true, x27, x26, x28, x29)≥FIBITER(x27, s(x26), x29, plus(x28, x29))) ⇒ IF(true, s(x27), s(x26), x14, x15)≥FIBITER(s(x27), s(s(x26)), x15, plus(x14, x15)))

We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
(7)    (IF(true, s(x24), 0, x14, x15)≥FIBITER(s(x24), s(0), x15, plus(x14, x15)))

We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x28,x29:lt(x26, x27)=true ⇒ IF(true, x27, x26, x28, x29)≥FIBITER(x27, s(x26), x29, plus(x28, x29))) with σ = [x28 / x14, x29 / x15] which results in the following new constraint:
(8)    (IF(true, x27, x26, x14, x15)≥FIBITER(x27, s(x26), x15, plus(x14, x15)) ⇒ IF(true, s(x27), s(x26), x14, x15)≥FIBITER(s(x27), s(s(x26)), x15, plus(x14, x15)))

To summarize, we get the following constraints P_≥ for the following pairs.

FIBITER(b, c, x, y) → IF(lt(c, b), b, c, x, y)
- (FIBITER(x4, s(x5), x7, x11)≥IF(lt(s(x5), x4), x4, s(x5), x7, x11))
IF(true, b, c, x, y) → FIBITER(b, s(c), y, plus(x, y))
- (IF(true, s(x24), 0, x14, x15)≥FIBITER(s(x24), s(0), x15, plus(x14, x15)))
- (IF(true, x27, x26, x14, x15)≥FIBITER(x27, s(x26), x15, plus(x14, x15)) ⇒ IF(true, s(x27), s(x26), x14, x15)≥FIBITER(s(x27), s(s(x26)), x15, plus(x14, x15)))

The constraints for P_> respective P_bound 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 P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [18]:

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

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 P_bound:

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

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.