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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
  ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 QDPOrderProof (⇔)
    ↳26 QDP
    ↳27 PisEmptyProof (⇔)
    ↳28 TRUE
  ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, y)
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(minus(x, y), z) → PLUS(y, z)
APP(cons(x, l), k) → APP(l, k)
SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))
SUM(cons(x, cons(y, l))) → PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))
SUM(app(l, cons(x, cons(y, k)))) → APP(l, sum(cons(x, cons(y, k))))
SUM(app(l, cons(x, cons(y, k)))) → SUM(cons(x, cons(y, k)))
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(s(x), s(y)) → IF(gt(x, y), x, y)
PLUS(s(x), s(y)) → GT(x, y)
PLUS(s(x), s(y)) → IF(not(gt(x, y)), id(x), id(y))
PLUS(s(x), s(y)) → NOT(gt(x, y))
PLUS(s(x), s(y)) → ID(x)
PLUS(s(x), s(y)) → ID(y)
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), x) → IF(gt(x, x), id(x), id(x))
PLUS(s(x), x) → GT(x, x)
PLUS(s(x), x) → ID(x)
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(id(x), s(y)) → IF(gt(s(y), y), y, s(y))
PLUS(id(x), s(y)) → GT(s(y), y)
NOT(x) → IF(x, false, true)
GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

APP(cons(x, l), k) → APP(l, k)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(cons(x, l), k) → APP(l, k)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[APP2, cons2]

Status:
cons2: [2,1]
APP2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(16) Obligation:

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

SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
cons(x1, x2)  =  cons(x1, x2)
plus(x1, x2)  =  plus(x2)
if(x1, x2, x3)  =  if
false  =  false
not(x1)  =  not(x1)
true  =  true
id(x1)  =  id(x1)
zero  =  zero
s(x1)  =  s
gt(x1, x2)  =  gt
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
SUM1 > cons2 > [plus1, id1, s]
if > [plus1, id1, s]
[false, gt] > true > [plus1, id1, s]
not1 > [plus1, id1, s]
zero > [plus1, id1, s]
0 > [plus1, id1, s]

Status:
gt: []
if: []
true: []
s: []
0: []
cons2: [2,1]
zero: []
SUM1: [1]
plus1: [1]
not1: [1]
false: []
id1: [1]


The following usable rules [FROCOS05] were oriented:

plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(s(x), y) → s(plus(x, y))
plus(0, y) → y

(18) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(22) Obligation:

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

MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
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.


MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  x1
minus(x1, x2)  =  x1
plus(x1, x2)  =  plus
s(x1)  =  s(x1)
if(x1, x2, x3)  =  if(x3)
false  =  false
not(x1)  =  x1
true  =  true
id(x1)  =  id(x1)
zero  =  zero
gt(x1, x2)  =  gt
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > id1 > plus
if1 > plus
false > plus
zero > [true, gt] > plus
0 > plus

Status:
zero: []
gt: []
if1: [1]
plus: []
true: []
false: []
s1: [1]
id1: [1]
0: []


The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

MINUS(minus(x, y), z) → MINUS(x, plus(y, z))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x1, x2)
minus(x1, x2)  =  minus(x1)
plus(x1, x2)  =  plus(x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
false  =  false
not(x1)  =  not
true  =  true
id(x1)  =  id(x1)
zero  =  zero
s(x1)  =  s
gt(x1, x2)  =  gt(x1, x2)
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
minus1 > MINUS2 > [plus1, id1, s]
not > if3 > [plus1, id1, s]
not > [false, zero] > [plus1, id1, s]
not > true > [plus1, id1, s]
gt2 > [false, zero] > [plus1, id1, s]
0 > [plus1, id1, s]

Status:
true: []
not: []
s: []
gt2: [1,2]
0: []
if3: [2,1,3]
zero: []
minus1: [1]
plus1: [1]
MINUS2: [1,2]
false: []
id1: [1]


The following usable rules [FROCOS05] were oriented:

plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(s(x), y) → s(plus(x, y))
plus(0, y) → y

(26) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
if(x1, x2, x3)  =  if(x1, x2, x3)
false  =  false
not(x1)  =  not
true  =  true
id(x1)  =  id(x1)
plus(x1, x2)  =  plus
zero  =  zero
gt(x1, x2)  =  gt(x1, x2)
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
QUOT2 > s1 > minus1
[if3, plus, gt2] > not > minus1
[if3, plus, gt2] > id1 > s1 > minus1
false > minus1
true > minus1
zero > minus1
0 > minus1

Status:
QUOT2: [2,1]
true: []
plus: []
not: []
gt2: [2,1]
0: []
if3: [1,3,2]
zero: []
minus1: [1]
false: []
s1: [1]
id1: [1]


The following usable rules [FROCOS05] were oriented:

minus(s(x), s(y)) → minus(x, y)
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(x, 0) → x

(31) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE