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Which of the following propositional statements is TRUE ?

A) ∀x ∀z ∃y [ P(x,y) ]---> ∃y ∀x ∀z [ P(x,y) ]

B) ∃y ∀x ∀z [ P(x,y) ]---> ∀x ∀z ∃y [ P(x,y) ]

I think in this

A) False 

B) True

Explanation:-

Domain :- Set of all negative numbers. {-inf to -1}

P(x,y) --> x>=y;

A)  ∀x ∀z ∃y [ P(x,y) ]---> ∃y ∀x ∀z [ P(x,y) ]

Solving the LHS part :-

∀x ∀z ∃y [ P(x,y) ]

In this first we have to set the value of x then moving towards the value of y.

x = all value should satisfy the condition

y = x-1;

if x = -3, y = -4 Condition satisfied

if x = -1, y = -2 Condition satisfied.

As we can see for all value of x there exist a value y for which the condition is always holds true. So, LHS part is True.

Moreover, LHS part always remain true because first we are setting the value of x, and then after setting x we can set y according always. hence it will always true.

Solving the RHS part :-

∃y ∀x ∀z [ P(x,y) ]

In this first we have to set the value of y then moving toward the value of x

y = -1, (Setting the value of y to -1)

x = all value should satisfy the condition

if y = -1, x = -1 Condition satisfied

if y = -1, x = -2 Condition not satisfied,

Since, The condition is not true for all value of x, Hence we can say that RHS part is false 

Conclusion:-

By the rule of implication, True -> False = False

Hence, part A will be False,

B) ∃y ∀x ∀z [ P(x,y) ]---> ∀x ∀z ∃y [ P(x,y) ]

From the above explanation we can conclude that LHS part of this equation is false and RHS part is True.

By the rule of implication, False -> True = True

Finally conclude that:

A) False

B) True.

Let me know where it is going wrong.

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