A system modeled as lossless compressed system is not wrong

But what about a system modeled as a (lossless) compressed system. This model is not wrong. Does not that disprove ""Essentially, ALL models are wrong,...". The said model is not wrong and is useful also.
So is there any fallacy in my understanding ?

• I'd argue that it's not a model at all. – Glen_b -Reinstate Monica Aug 28 '17 at 12:40
• @Glen_b If it is not a model then what it is ? (many lossless compression techniques rely on statistical Models) – gpuguy Aug 28 '17 at 14:41
• A lossless compression technique is exactly that -- a compression technique. A model (necessarily) abstracts away some unimportant details (unimportant to the particular purpose at least; they may be important to other purposes). Viewed through the less of compression, a model is deliberately lossy. (Using statistical models is different from being a statistical model.) – Glen_b -Reinstate Monica Aug 28 '17 at 14:53
• @Glen_b Ok. Can we consider lossy compressed system as a model ? Because a lossy compression is normally a result of removing redundancy (i.e unimportant details). – gpuguy Aug 28 '17 at 16:28
• You might in some circumstances look at it that way; if the loss was almost all of unimportant detail then you could perhaps argue for regarding it as a model. For it to be a statistical model you'd need to include some probabilistic aspect in the model relating observation to underlying "signal" – Glen_b -Reinstate Monica Aug 29 '17 at 1:19

I'd argue that it's not a model at all. A lossless compression technique is exactly that -- a compression technique. A model (necessarily) abstracts away some unimportant details (unimportant to the particular purpose at least; they may be important to other purposes). Viewed through the less of compression, a model is deliberately lossy. (Using statistical models is different from being a statistical model.)

– Glen_b