This is trivial by Newton's second law, $F=ma$, at the center of mass.

No, that acceleration includes the rotational one.

@hellofriends I think you have it backwards in the title to your post. If the net force on an object is zero, then there will be no linear acceleration.

@VincentThacker you would have to prove that the integral of rotational acceleration ($a \times r + \omega \times \omega \times r)$ along the body is zero

@BobD angular aceleration also translates as acceleration. If net force is zero implies total acceleration is zero, then it is enough to garantee that the body won't have angular acceleration. But it only provides zero linear acceleration specically.

Don't confuse net force with net torque.

@ACB I know what the equations are suposed to mean, can you prove what they mean mathematically in a random rigid body?

I'm not confusing net force with net torque. Net force results in linear acceleration of the COM. Net force not acting thru the COM additionally results in angular acceleration. A couple (two equal, opposite and parallel forces) results in pure rotation and no linear acceleration of the COM since the net force is zero.

@hellofriends You know what you think the equations are supposed to mean, but I'm not sure that you think what I think they mean. You really need to define what you mean by "zero linear acceleration" here. It is obvious that the linear acceleration is different at different points on a rotating rigid body.

@BobD so Newton's second law should be changed for: Force is equal mass times linear acceleration. And thus, we could have spining bodies without an external force acting on them, since centripetal acceleration is just an independent phenomena causes by an unrelated property called torque.

@alephzero if the linear acceleration is different in different poins of a rigid body, it is not a rigid body. Particles must keep their relative position with respect to each other.

@hellofriends The statement of Newton's second law applies to point particles which have no rotation by definition of what "point particle" means. In general, rotating rigid bodies have non-zero, but equal and opposite, force pairs between every pair of particles in the body, but the net force on each particle is not zero. That is why reason I'm not sure what you mean by "net force of zero."

@alephzero In general, rotating rigid bodies have non-zero, but equal and opposite, force pairs between every pair of particles in the body, but the net force on each particle is not zero. Only if angular acceleration is non-zero.

@hellofriends IMO this won't be resolved until you stop describing everything as a "word problem" and start doing some math.

@hellofriends It seems that you have some serious confusion on Newton's second law and torque about center of mass, but without clarifying your problem, it is impossible to address.

@Gert I would rather change "force pairs" in my comment to "internal force pairs" than bring in nonlinear angular accelerations to confuse the OP even more - but you are right of course.

@alephzero you are correct, net force on points are not zero. When we do the equilibrium of a rigid body, we use 6 equations: net force ON THE BODY equals zero and net torque equals zero. Why the first one garantess the body will transfer in constant linear speed?

@VincentThacker you are the one confused. I know every movement of a rigid body can be decomposed in translation and rotation. I'm just wondering why doing the net force of the body as a whole be zero causes It's translation acceleration to be zero.

@hellofriends Refer to my first comment. As long as the net force is zero, the COM will not accelerate, and this is trivial by definition of $F=ma$, because $F$ is the net force and $a$ is the acceleration of the COM.

@VincentThacker this by it's own doesn't prove it can't be rotating around some axis, you would need to prove that the center of mass only have translational acceleration in this case

@hellofriends Again, this is trivial, because the COM is defined to be the point that accelerates linearly with $\vec{a}$ whenever a force of $m\vec{a}$ is applied anywhere on the body. So there is nothing to prove. What you are asking is equivalent to first defining, say, $y=2x$, and then asking why $x=y/2$.

@VincentThacker all I want is a mathematical proof of that property for the center of mass. It rejects any angular kind of acceleration on a net force of zero. This is not just a definition.

@hellofriends "this by it's own doesn't prove it can't be rotating around some axis". I did not say that it won't rotate. I am just saying that the COM will not accelerate if the net force is zero. If the forces have a net torque, then the mass will, of course, experience angular acceleration. But the COM will not accelerate.

@VincentThacker but a net force of zero will cancel translational accelerational and I need a reason for this. This is not trivial. The fact it will have zero total acceleration is fine for me, I need a proof for the translational part.

For a rigid body, the mass density $f(\vec{x})$ can only evolve to $f(R(\vec{x}-\vec{c}))$ with $R^T=R^{-1}$. It sounds like you want to prove zero net force implies $\ddot{\vec{c}}=\vec{0},\,\ddot{R}=0$. Computing the second time derivative of the COM's position handles the first part, but the second part probably also requires zero net torque. Anyway, that kind of formalism is where a rigorous proof would have to start.

@hellofriends I think that there is little point in continuing this discussion, because you seem too convinced that you are right to listen to what others are saying. In any case, you should have a look at physics.stackexchange.com/q/118461

@hellofriends [...] and I need a reason for this. This is not trivial. It IS trivial. Not to mention seriously boring...

@Gert if it is trivial, please prove that in a system of three particles whose distance between each other is fixed will not tranlsadate with respect to the ground if one apply a force one $1N$ on one of them and a force of $1N$ in the opposite direction on another.

@hellofriends In the case you provided, the particles will move accordingly, but the center of mass of the whole system will not moves.

@Kksen Yes, show me that the particles will only spin and not transladate. I don't care about the center of mass unless you can prove me that the center of mass can't spin when the net force is zero.

You mean when net force is zero, the center of mass is possible to spin?

@Kksen I want a prove that this can't be the case. If it could, then the center of mass would be able to have a translational acceleration that would cancel with the rotational one. But apperently, when a rigid body has a net force of zero, the translational acceleration of all points in the body is zero.

What I want to know is do the proposed answers solve your doubts? Seems like the one by @Miechael Seifert addressed it on point.