- #1

solakis1

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For all A,B we define the floor value of A denoted by [A] to be an iteger B such that : $[A]=B\Leftrightarrow B\leq A<B+1$

And in symbols $\forall A\forall B ( [A]=B\Leftrightarrow B\leq A<B+1\wedge B\in Z)$,then prove:

For all A $ [A]\leq A<[A]+1$