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  • Set of $LL(1)$ grammars is a proper subset of the set of $LR(1)$ grammars.
  • $\color{magenta} \epsilon$ free  $LL(1)$ grammars are $SLR(1)$ grammars. So they are $LALR(1)$ too.
  • Now among those grammars that contain $\epsilon$ productions, If the symbols of that grammar have both empty as well as non-empty derivations, then that grammar is $LALR(1)$ too.
  • But if any symbol contain only empty derivation then that grammar may or may not $LALR(1).$

So set of $LL(1)$ grammars is not subset of the set of $LALR(1)$ grammars. 
 

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