Termination w.r.t. Q proof of /tmp/tmpE40-8l/cade16.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:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

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:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
LENGTH(cons(x, l)) → LENGTH(l)
COND(false, n, l) → NTHTAIL(s(n), l)
NTHTAIL(n, l) → GE(n, length(l))
COND(false, n, l) → TAIL(nthtail(s(n), l))
GE(s(u), s(v)) → GE(u, v)
NTHTAIL(n, l) → LENGTH(l)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
LENGTH(cons(x, l)) → LENGTH(l)
COND(false, n, l) → NTHTAIL(s(n), l)
NTHTAIL(n, l) → GE(n, length(l))
COND(false, n, l) → TAIL(nthtail(s(n), l))
GE(s(u), s(v)) → GE(u, v)
NTHTAIL(n, l) → LENGTH(l)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

GE(s(u), s(v)) → GE(u, v)

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

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

↳ 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:

GE(s(u), s(v)) → GE(u, v)

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:

GE(s(u), s(v)) → GE(u, v)
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:

LENGTH(cons(x, l)) → LENGTH(l)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

↳ 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:

LENGTH(cons(x, l)) → LENGTH(l)

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

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

↳ 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:

LENGTH(cons(x, l)) → LENGTH(l)

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

LENGTH(cons(x, l)) → LENGTH(l)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
COND(false, n, l) → NTHTAIL(s(n), l)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

↳ 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:

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
COND(false, n, l) → NTHTAIL(s(n), l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))

↳ 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:

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
COND(false, n, l) → NTHTAIL(s(n), l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(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 NTHTAIL(n, l) → COND(ge(n, length(l)), n, l) the following chains were created:

We consider the chain COND(false, n, l) → NTHTAIL(s(n), l), NTHTAIL(n, l) → COND(ge(n, length(l)), n, l) which results in the following constraint:
(1) (NTHTAIL(s(x2), x3)=NTHTAIL(x4, x5) ⇒ NTHTAIL(x4, x5)≥COND(ge(x4, length(x5)), x4, x5))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (NTHTAIL(s(x2), x3)≥COND(ge(s(x2), length(x3)), s(x2), x3))

For Pair COND(false, n, l) → NTHTAIL(s(n), l) the following chains were created:

We consider the chain NTHTAIL(n, l) → COND(ge(n, length(l)), n, l), COND(false, n, l) → NTHTAIL(s(n), l) which results in the following constraint:
(3)    (COND(ge(x6, length(x7)), x6, x7)=COND(false, x8, x9) ⇒ COND(false, x8, x9)≥NTHTAIL(s(x8), x9))

We simplified constraint (3) using rules (I), (II), (III), (VII) which results in the following new constraint:
(4)    (length(x7)=x12∧ge(x6, x12)=false ⇒ COND(false, x6, x7)≥NTHTAIL(s(x6), x7))

We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on ge(x6, x12)=false which results in the following new constraints:
(5)    (false=false∧length(x7)=s(x14) ⇒ COND(false, 0, x7)≥NTHTAIL(s(0), x7))

(6)    (ge(x15, x16)=false∧length(x7)=s(x16)∧(∀x17:ge(x15, x16)=false∧length(x17)=x16 ⇒ COND(false, x15, x17)≥NTHTAIL(s(x15), x17)) ⇒ COND(false, s(x15), x7)≥NTHTAIL(s(s(x15)), x7))

We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
(7)    (length(x7)=s(x14) ⇒ COND(false, 0, x7)≥NTHTAIL(s(0), x7))

We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on length(x7)=s(x16) which results in the following new constraint:
(8)    (s(length(x22))=s(x16)∧ge(x15, x16)=false∧(∀x17:ge(x15, x16)=false∧length(x17)=x16 ⇒ COND(false, x15, x17)≥NTHTAIL(s(x15), x17))∧(∀x23,x24,x25:length(x22)=s(x23)∧ge(x24, x23)=false∧(∀x25:ge(x24, x23)=false∧length(x25)=x23 ⇒ COND(false, x24, x25)≥NTHTAIL(s(x24), x25)) ⇒ COND(false, s(x24), x22)≥NTHTAIL(s(s(x24)), x22)) ⇒ COND(false, s(x15), cons(x21, x22))≥NTHTAIL(s(s(x15)), cons(x21, x22)))

We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on length(x7)=s(x14) which results in the following new constraint:
(9)    (s(length(x19))=s(x14)∧(∀x20:length(x19)=s(x20) ⇒ COND(false, 0, x19)≥NTHTAIL(s(0), x19)) ⇒ COND(false, 0, cons(x18, x19))≥NTHTAIL(s(0), cons(x18, x19)))

We simplified constraint (9) using rules (I), (II), (IV) which results in the following new constraint:
(10)    (COND(false, 0, cons(x18, x19))≥NTHTAIL(s(0), cons(x18, x19)))

We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
(11)    (length(x22)=x16∧ge(x15, x16)=false∧(∀x17:ge(x15, x16)=false∧length(x17)=x16 ⇒ COND(false, x15, x17)≥NTHTAIL(s(x15), x17))∧(∀x23,x24,x25:length(x22)=s(x23)∧ge(x24, x23)=false∧(∀x25:ge(x24, x23)=false∧length(x25)=x23 ⇒ COND(false, x24, x25)≥NTHTAIL(s(x24), x25)) ⇒ COND(false, s(x24), x22)≥NTHTAIL(s(s(x24)), x22)) ⇒ COND(false, s(x15), cons(x21, x22))≥NTHTAIL(s(s(x15)), cons(x21, x22)))

We simplified constraint (11) using rule (VI) where we applied the induction hypothesis (∀x17:ge(x15, x16)=false∧length(x17)=x16 ⇒ COND(false, x15, x17)≥NTHTAIL(s(x15), x17)) with σ = [x17 / x22] which results in the following new constraint:
(12)    (COND(false, x15, x22)≥NTHTAIL(s(x15), x22)∧(∀x23,x24,x25:length(x22)=s(x23)∧ge(x24, x23)=false∧(∀x25:ge(x24, x23)=false∧length(x25)=x23 ⇒ COND(false, x24, x25)≥NTHTAIL(s(x24), x25)) ⇒ COND(false, s(x24), x22)≥NTHTAIL(s(s(x24)), x22)) ⇒ COND(false, s(x15), cons(x21, x22))≥NTHTAIL(s(s(x15)), cons(x21, x22)))

We simplified constraint (12) using rule (IV) which results in the following new constraint:
(13)    (COND(false, x15, x22)≥NTHTAIL(s(x15), x22) ⇒ COND(false, s(x15), cons(x21, x22))≥NTHTAIL(s(s(x15)), cons(x21, x22)))

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

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
- (NTHTAIL(s(x2), x3)≥COND(ge(s(x2), length(x3)), s(x2), x3))
COND(false, n, l) → NTHTAIL(s(n), l)
- (COND(false, 0, cons(x18, x19))≥NTHTAIL(s(0), cons(x18, x19)))
- (COND(false, x15, x22)≥NTHTAIL(s(x15), x22) ⇒ COND(false, s(x15), cons(x21, x22))≥NTHTAIL(s(s(x15)), cons(x21, x22)))

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(COND(x₁, x₂, x₃)) = -1 - x₁ - x₂ + x₃
POL(NTHTAIL(x₁, x₂)) = -1 - x₁ + x₂
POL(c) = -2
POL(cons(x₁, x₂)) = 1 + x₂
POL(false) = 0
POL(ge(x₁, x₂)) = 0
POL(length(x₁)) = 0
POL(nil) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

COND(false, n, l) → NTHTAIL(s(n), l)

The following pairs are in P_bound:

COND(false, n, l) → NTHTAIL(s(n), l)

The following rules are usable:

false → ge(0, s(v))
ge(u, v) → ge(s(u), s(v))
true → ge(u, 0)

↳ 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:

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(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.