##### How to check if statements are equivalent or not?

we have to check whether the statement is equivalent or not. For that we try to find a true false case and if it exists then that will imply that it is not equivalent. Now for creating True in LHS we say that there exists an x such that either P(x) or Q(x) is true for it. for RHS can we say that there exists an x such that p(x) is false and there exists another x for which Q(x) is false.??????? Since we know that both the x in RHS are not similar and there is no condition in LHS for both P(x) and Q(x) to be true simultaneously and any one of them being true at a time can do the job. Thanks

x should be same in your discussion.

∃x [ P(x) ∨ Q(x) ] ⇔ ∃xP(x) ∨ ∃xQ(x)

To prove this we have to check two cases:

1. ∃x [ P(x) ∨ Q(x) ] → ∃xP(x) ∨ ∃xQ(x)

2. ∃xP(x) ∨ ∃xQ(x) → ∃x [ P(x) ∨ Q(x) ]

If any one of the case is false, then equivalency can not be applied.

Take,

x be the domain of students in a school.

P(x) = x is in Mathematics department.

Q(x) = x is in Biology department.

Check case 1:∃x [ P(x) ∨ Q(x) ] → ∃xP(x) ∨ ∃xQ(x)

To make it false, we have to make LHS as True and RHS as False.

Here, LHS says, there is a student who is either in Mathematics department or in Biology department. If we will make this as True, we can not make RHS as False. So this case 1 is True.

Check case 2:∃xP(x) ∨ ∃xQ(x) → ∃x [ P(x) ∨ Q(x) ]

Here, if we will make LHS as True, then there is at least one student who is either in Mathematics or Biology department. Now, we can not make RHS as False. So case2 is True.

Hence, Formula is correct.