Termination w.r.t. Q proof of /tmp/tmpQGBLif/Ex26

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

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

(0) Obligation:

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

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

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:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

(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:

TERMS(N) → SQR(N)
TERMS(N) → TERMS(s(N))
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
SQR(s(X)) → DBL(X)
DBL(s(X)) → DBL(X)
ADD(s(X), Y) → ADD(X, Y)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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.

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2) = x1
s(x1) = s(x1)
cons(x1, x2) = cons(x1)
terms(x1) = terms(x1)
recip(x1) = x1
sqr(x1) = sqr(x1)
0 = 0
add(x1, x2) = add(x1, x2)
dbl(x1) = dbl(x1)
first(x1, x2) = x2
nil = nil

Lexicographic path order with status [LPO].
Precedence:

terms₁ > cons₁ > nil
terms₁ > sqr₁ > add₂ > s₁ > nil
terms₁ > sqr₁ > dbl₁ > s₁ > nil
terms₁ > sqr₁ > dbl₁ > 0 > nil

Status:

s₁: [1]
cons₁: [1]
terms₁: [1]
sqr₁: [1]
0: []
add₂: [2,1]
dbl₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

(9) Obligation:

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

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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:

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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.

ADD(s(X), Y) → ADD(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = ADD(x1, x2)
s(x1) = s(x1)
terms(x1) = terms(x1)
cons(x1, x2) = cons
recip(x1) = x1
sqr(x1) = sqr(x1)
0 = 0
add(x1, x2) = add(x1, x2)
dbl(x1) = dbl(x1)
first(x1, x2) = x2
nil = nil

Lexicographic path order with status [LPO].
Precedence:

ADD₂ > nil
terms₁ > cons > nil
terms₁ > sqr₁ > add₂ > s₁ > nil
terms₁ > sqr₁ > dbl₁ > s₁ > nil
terms₁ > sqr₁ > dbl₁ > 0 > nil

Status:

ADD₂: [1,2]
s₁: [1]
terms₁: [1]
cons: []
sqr₁: [1]
0: []
add₂: [1,2]
dbl₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

(14) Obligation:

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

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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:

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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.

DBL(s(X)) → DBL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL(x1) = DBL(x1)
s(x1) = s(x1)
terms(x1) = terms
cons(x1, x2) = cons
recip(x1) = recip
sqr(x1) = sqr(x1)
0 = 0
add(x1, x2) = add(x1, x2)
dbl(x1) = dbl(x1)
first(x1, x2) = x2
nil = nil

Lexicographic path order with status [LPO].
Precedence:

DBL₁ > nil
terms > cons > nil
terms > recip > nil
terms > sqr₁ > add₂ > s₁ > nil
terms > sqr₁ > dbl₁ > s₁ > nil
terms > sqr₁ > dbl₁ > 0 > nil

Status:

DBL₁: [1]
s₁: [1]
terms: []
cons: []
recip: []
sqr₁: [1]
0: []
add₂: [2,1]
dbl₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

(19) Obligation:

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

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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:

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SQR(s(X)) → SQR(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1) = SQR(x1)
s(x1) = s(x1)
terms(x1) = terms
cons(x1, x2) = cons
recip(x1) = recip
sqr(x1) = sqr(x1)
0 = 0
add(x1, x2) = add(x1, x2)
dbl(x1) = dbl(x1)
first(x1, x2) = x2
nil = nil

Lexicographic path order with status [LPO].
Precedence:

SQR₁ > nil
terms > cons > nil
terms > recip > nil
terms > sqr₁ > add₂ > s₁ > nil
terms > sqr₁ > dbl₁ > s₁ > nil
terms > sqr₁ > dbl₁ > 0 > nil

Status:

SQR₁: [1]
s₁: [1]
terms: []
cons: []
recip: []
sqr₁: [1]
0: []
add₂: [2,1]
dbl₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

(24) Obligation:

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

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

TERMS(N) → TERMS(s(N))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))

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