# how is this method of proving shift invariance correct

I got this answer from here. But i don't know how this method is correct. Until now to prove shift invariance what i did was :
step 1: delay output and write it in terms of input.
step 2: delay input and write the modified output
step 3: if output from step 1 and 2 are same then system is shift invariant.

what this guy has done is "different". the oppenheim book and proakis manokalis book don't follow this method. which book follows this method?

the method.

The test for this is $$x_{1}[n]=x[n-n_{0}]$$ $$y_{1}[n]=y[n-n_{0}]$$

So let's first express in terms of just $$y[n]$$. $$y[n]=\frac{x[n-3]-y[n-3]-y[n-1]}{x[n]}$$ Next, we go through the test. $$y[n-n_{0}]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]}$$ Substituting $$y_{1}[n]$$ $$y_{1}[n]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]}$$ $$y_{1}[n]=\frac{x[n-3-n_{0}]-y_{1}[n-3]-y_{1}[n-1]}{x[n-n_{0}]}$$ And finally substituting $$x_{1}[n]$$ $$y_{1}[n]=\frac{x_{1}[n-3]-y_{1}[n-3]-y_{1}[n-1]}{x_{1}[n]}$$ Because the shifted sequence has the exact same relationship, it is said to be time-invariant. Most systems that don't alter $$n$$ or $$t$$ meet this. Be wary of anything multiplying or otherwise messing with the vector arguments beyond simple delays.

The method relies on the fact that the relationship between the output $$y[n]$$ and input $$x[n]$$ remains the same between $$y[n-d]$$ and $$x[n-d]$$.
As an absurd example, think of the following. Consider the equation of a circle at origin: $$x^2 + y^2 = r^2$$. Now, shift this circle to the point $$(a,b)$$ and re-write its equation: $$(x-a)^2 + (y-b)^2 = r^2$$. Finally, redefine the variables $$x' = x-a$$ and $$y' = y-b$$ and rewrite the equation: $$(x')^2 + (y')^2 = r^2$$.