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:

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

double₁ > 0 > s₁
half₁ > s₁
-₂ > s₁
if₃ > s₁

Status:

double₁: [1]
0: multiset
s₁: multiset
half₁: [1]
-₂: [1,2]
if₃: multiset

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

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → 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