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 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE

(0) Obligation:

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(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:

APP(x, y) → HELPA(0, plus(length(x), length(y)), x, y)
APP(x, y) → PLUS(length(x), length(y))
APP(x, y) → LENGTH(x)
APP(x, y) → LENGTH(y)
PLUS(x, s(y)) → PLUS(x, y)
LENGTH(cons(x, y)) → LENGTH(y)
HELPA(c, l, ys, zs) → IF(ge(c, l), c, l, ys, zs)
HELPA(c, l, ys, zs) → GE(c, l)
GE(s(x), s(y)) → GE(x, y)
IF(false, c, l, ys, zs) → HELPB(c, l, ys, zs)
TAKE(s(c), cons(x, xs), ys) → TAKE(c, xs, ys)
TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)
HELPB(c, l, ys, zs) → TAKE(c, ys, zs)
HELPB(c, l, ys, zs) → HELPA(s(c), l, ys, zs)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(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 5 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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.


TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2, x3)  =  TAKE(x3)
s(x1)  =  s
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
nil > [TAKE1, s]
cons2 > [TAKE1, s]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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.


GE(s(x), s(y)) → GE(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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:

IF(false, c, l, ys, zs) → HELPB(c, l, ys, zs)
HELPB(c, l, ys, zs) → HELPA(s(c), l, ys, zs)
HELPA(c, l, ys, zs) → IF(ge(c, l), c, l, ys, zs)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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

(18) Obligation:

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

LENGTH(cons(x, y)) → LENGTH(y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


LENGTH(cons(x, y)) → LENGTH(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(x, s(y)) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
s1 > PLUS1


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

app(x0, x1)
plus(x0, 0)
plus(x0, s(x1))
length(nil)
length(cons(x0, x1))
helpa(x0, x1, x2, x3)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
take(0, cons(x0, xs), x1)
take(0, nil, cons(x0, x1))
take(s(x0), cons(x1, xs), x2)
take(s(x0), nil, cons(x1, x2))
helpb(x0, x1, x2, x3)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE