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'A' is set of all possible schedules
'C' is set of all possible schedules that are guaranteed to produce a correct final result
'S' is the set of all serializable schedules
'P' is the set of all schedules possible under 2-phase locking protocol

Which is FALSE?

(A) P⊆C
(B) S⊂P
(C) S⊆P
(D) P⊂C

Please provide reason for the answer. (Given answer is A)

in Databases by Active (2.2k points) | 79 views
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S includes view serialisable as well as conflict serialisable, then how can S is a subset of P correct, when P is containing all conflict serialisable schedule (as 2PL guarantees conflict serialisability)..., when we know that #view serialisable schedules are greater than equal to #conflict serialisable schedule.
0
where  you find difficulty here..it's simple set theory logic if you already know the concept of conflict serializability , 2 phase locking system

+ you didn't post the given answer with the question , until and unless someone  ask you because it just spoil the spirit to do the question
0
how serializable schedule is proper subset of 2PL

is a correct option?
+1
yeah option B is also false
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could you please give your explanation for the answer.

According to me both option $B$ and $C$ are wrong as $2PL$ schedules are small schedules under all serializable schedules. Similarly option $A$ and $D$ are true bcz either set of all $2PL$ schedules produce correct result or some of them.

Then how option $A$ is answer?

 

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