a b c d

a

for every $L_1∈L$ $L_1⊥L_1$ (Not possible) (Not reflexive) $(L_1,L_2)∈R⇒L_1⊥L_2⇒(L_2,L_1)∈R$  (Symmetric) $(L_1,L_2)∈R,(L_2,L_3)∈R$ $L_1⊥L_2$ and $L_2⊥L_3⇒L_1||L_3$  (Not transitive)