Termination w.r.t. Q proof of /tmp/tmpgQnait/10.trs

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 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 PisEmptyProof (⇔)

    ↳24 TRUE

  ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 PisEmptyProof (⇔)

    ↳34 TRUE

  ↳35 QDP

(0) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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:

+¹(s(x), y) → +¹(x, y)
++¹(:(x, xs), ys) → ++¹(xs, ys)
SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))
SUM(:(x, :(y, xs))) → +¹(x, y)
SUM(++(xs, :(x, :(y, ys)))) → SUM(++(xs, sum(:(x, :(y, ys)))))
SUM(++(xs, :(x, :(y, ys)))) → ++¹(xs, sum(:(x, :(y, ys))))
SUM(++(xs, :(x, :(y, ys)))) → SUM(:(x, :(y, ys)))
-¹(s(x), s(y)) → -¹(x, y)
QUOT(s(x), s(y)) → QUOT(-(x, y), s(y))
QUOT(s(x), s(y)) → -¹(x, y)
LENGTH(:(x, xs)) → LENGTH(xs)
AVG(xs) → QUOT(hd(sum(xs)), length(xs))
AVG(xs) → HD(sum(xs))
AVG(xs) → SUM(xs)
AVG(xs) → LENGTH(xs)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

LENGTH(:(x, xs)) → LENGTH(xs)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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.

LENGTH(:(x, xs)) → LENGTH(xs)

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

(7) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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:

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2) = x1
s(x1) = s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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:

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

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(16) 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(-(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)
-(x1, x2) = -(x1)
0 = 0

Recursive Path Order [RPO].
Precedence:

s₁ > [QUOT₂, -₁]

The following usable rules [FROCOS05] were oriented:

-(s(x), s(y)) → -(x, y)
-(0, s(y)) → 0
-(x, 0) → x

(17) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

++¹(:(x, xs), ys) → ++¹(xs, ys)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

++¹(:(x, xs), ys) → ++¹(xs, ys)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2) = x1
:(x1, x2) = :(x1, x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) Obligation:

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(s(x), y) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x1)
s(x1) = s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > +^1₁

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) Obligation:

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

SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1) = SUM(x1)
:(x1, x2) = :(x2)
+(x1, x2) = +
s(x1) = s
0 = 0

Recursive Path Order [RPO].
Precedence:

SUM₁ > [:₁, +, s]

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) Obligation:

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

SUM(++(xs, :(x, :(y, ys)))) → SUM(++(xs, sum(:(x, :(y, ys)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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