every $\epsilon$-free LL(1) grammar is SLR(1) grammar means the set of $\epsilon$-free LL(1) grammars is within the set of SLR(1) grammars.

To prove 2 sets A and B are equal , we have to prove that $A\subseteq B$ and $B\subseteq A$. So, if we consider $\epsilon$-free LL(1) = SLR(1) is true then all the SLR(1) grammars should also be the subset of set of $\epsilon$-free LL(1) grammars which is not true.