This isn't some kind of EULA situation where a provision in the contract will be thrown out: the cameras are stated to be for non-commercial use because no one paid for a commercial license. Suing would be quick and dirty, and the user would be at fault.
No. There is no provision in the contract to buy a camera stating that it is for personal use only, unless it's clearly marked on the packaging. This means the license could only bind the user if the patents are valid. If they are found to be invalid, there is no consideration and so the license cannot bind the user to anything. If the patents are valid, the user might still be found not to be subject to the license (since they would have to agree to it, and most users aren't even aware of the license's existence - they could still be found to agree without knowledge if there was significant notices attached to the camera that most people would notice stating it could only be used if you agree to the license, but to my knowledge no camera is like that), but then they would be at fault for patent infringement, which generally does not require intent.
Would be tough... Patents are a property right. We don't normally take peoples' property away if they're not using it in ways that we'd like, but maybe you could make an argument under Kelo v. City of New London that it's an eminent domain taking. Plus, it wouldn't be invalidating the patent - it would be assigning it to the government, who then releases it free into the public domain.
Some patent laws provide provisions by which a patent holder can be seen to be abusing their patent and have conditions forced upon them, but these seem to get rarely used (Here's Canada's Patent Act as an example).
I seem to have heard this argument before. The Apollo fire. The loss of the Challenger. Repairs to the Hubble.
The difference here is that they know what the current safe maximum is, so they'll operate it for a while, tear the thing down and replace them with interconnects with higher safe maximums, and start it up again. They are aware that the lower safety factor must also mean a lower-power experiement (not that putting 7 TeV of energy into a single proton is normally considered low power).
Divide both sides by 0: (0/0)/0 = (1/1)/0 *** By associativity of division across the reals: 0/(0/0) = (1/1)/0
Uh, problem. Division isn't associative:
(1/2)/3 = 1/6
1/(2/3) = 3/2
Also, in the reals (which is your stated field), x/0 is taken to be undefined, meaning that your division of both sides by 0 isn't valid unless you define that too (which can be done, but you didn't do it).
The possession of a book becomes a substitute for reading it. -- Anthony Burgess