Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

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:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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:

SQR(s(x)) → +¹(sqr(x), s(double(x)))
SQR(s(x)) → SQR(x)
SQR(s(x)) → DOUBLE(x)
DOUBLE(s(x)) → DOUBLE(x)
+¹(x, s(y)) → +¹(x, y)
SQR(s(x)) → +¹(sqr(x), double(x))

The TRS R consists of the following rules:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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.

+¹(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)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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:

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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.

DOUBLE(s(x)) → DOUBLE(x)

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

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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:

SQR(s(x)) → SQR(x)

The TRS R consists of the following rules:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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.

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)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

The set Q consists of the following terms:

sqr(0)
sqr(s(x0))
double(0)
double(s(x0))
+(x0, 0)
+(x0, s(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