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

(0) Obligation:

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(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:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

(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:

GE(s(x), s(y)) → GE(x, y)
REV(x) → IF(x, eq(0, length(x)), nil, 0, length(x))
REV(x) → LENGTH(x)
IF(x, false, z, c, l) → HELP(s(c), l, x, z)
HELP(c, l, cons(x, y), z) → IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
HELP(c, l, cons(x, y), z) → APPEND(y, cons(x, nil))
HELP(c, l, cons(x, y), z) → GE(c, l)
APPEND(cons(x, y), z) → APPEND(y, z)
LENGTH(cons(x, y)) → LENGTH(y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) 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:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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.


LENGTH(cons(x, y)) → LENGTH(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
cons2 > LENGTH1

Status:
LENGTH1: [1]
cons2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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:

APPEND(cons(x, y), z) → APPEND(y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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.


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

Lexicographic path order with status [LPO].
Precedence:
cons2 > APPEND1

Status:
APPEND1: [1]
cons2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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.


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)  =  GE(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
s1 > GE1

Status:
GE1: [1]
s1: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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:

HELP(c, l, cons(x, y), z) → IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
IF(x, false, z, c, l) → HELP(s(c), l, x, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
rev(x0)
if(x0, true, x1, x2, x3)
if(x0, false, x1, x2, x3)
help(x0, x1, cons(x2, x3), x4)
append(nil, x0)
append(cons(x0, x1), x2)
length(nil)
length(cons(x0, x1))

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