p ~(Op & (~p)) is equivalent to
p Op
p. Q.E.D.
Second proof: We already concluded that the most preferred point (Op) would also be the chosen point (p). Thus
p Op
p. (If the point is not preferred, then the implication is true ex vacuoso.) Q.E.D.
p ~(Op & (~p)) is equivalent to
p Op
p. Q.E.D.
Second proof: We already concluded that the most preferred point (Op) would also be the chosen point (p). Thus
p Op
p. (If the point is not preferred, then the implication is true ex vacuoso.) Q.E.D.