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) QTRSRRRProof (EQUIVALENT transformation)

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

sqr₁ > 0 > s₁
sqr₁ > +₂ > s₁
sqr₁ > double₁ > s₁

Status:

sqr₁: [1]
s₁: [1]
double₁: [1]
0: []
+₂: [2,1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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)))

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE