case1: ∀x(P(x) ∧ Q(x)) =. ∀xP(x) ∧ ∀xQ(x)
assume a set S={6,12,18,24}
let p(x) = "x divisible by 2 "
q(x) = "x divisible by 3 "
LHS
∀x(P(x) ∧ Q(x)) is true always for above p(x) & q(x)....as for every x ( "x divisible by 2 " & "x divisible by 3 ")
..RHS:::∀xP(x) ∧ ∀xQ(x) is also true ...as {for every x "x is divisible by 2" & for every x " x is divisible by 3"}
∀x(P(x) ∧ Q(x)) =. ∀xP(x) ∧ ∀xQ(x)
T = T
as LHS & RHS are equal
so we can distribute....a universal quantifier over a conjunction
case 2:∀x(P(x) v Q(x)) = ∀xP(x) v ∀xQ(x)
let set S={2,6,4,8,9}
let p(x) = "x divisible by 2 "
q(x) = "x divisible by 3 "
LHS :∀x(P(x) v Q(x)) ...this is true..always bcoz for all x ("x divisible by 2 "OR "x divisible by 3 ")
∀xP(x) v ∀xQ(x)...this is false bcoz for all x ("x divisible by 2 ") OR for all x( "x divisible by 3 ")
counter example : "element 9 is divisible by 2 OR element 8 is divisible by 3" this is false
∀x(P(x) v Q(x)) = ∀xP(x) v ∀xQ(x)
T = F
as LHS & RHS are not equal
...hence ...a universal quantifier cannot be distributed over a disjunction.
case 3: ∃x(P(x) v Q(x)) = ∃xP(x) v ∃xQ(x)
lets set S={2,6,4,8,9}
let p(x) = "x divisible by 2 "
q(x) = "x divisible by 3 "
LHS::::∃x(P(x) v Q(x)...is true..boz there exist "x' for which ( "x divisible by 2 " OR "x divisible by 3 ")...u can take any element..lets takeEXP1- ===="9 divisible by 2 " OR "9 divisible by 3" ::: F V T = T
take EXP2- ::: "6 divisible by 2 " OR "6 divisible by 3" T V T = T
RHS ::::∃xP(x) v ∃xQ(x) ..is true bcoz {there exist "x' for which ( "x divisible by 2) " OR exist "x' for which "x divisible by 3 }")
let take EXP1- === "9 divisible by 2 " OR "6 divisible by 3" F V T = T
let takeEXP2 - ==="2 divisible by 2 " OR "9 divisible by 3" T V T = T
∃x(P(x) v Q(x)) =∃xP(x) v ∃xQ(x)
T ====== T
as LHS & RHS are equal
we can distribute an existential quantifier over a disjunction.
case4:∃x(P(x) ∧ Q(x)) = ∃xP(x) ∧ ∃xQ(x)
assume a set S={2,4,3,9}
let p(x) = "x divisible by 2 "
q(x) = "x divisible by 3 "
LHS :::∃x(P(x) ∧ Q(x)) is false always...as there exist no number which divisible by both 2 & 3..in above set
RHS ::∃xP(x) ∧∃xQ(x) is true always...as "there exist a number which divisible by 2 " and "there exist a number which divisible by 3 "
∃x(P(x) ∧ Q(x)) = ∃xP(x) ∧ ∃xQ(x)
F = T
as LHS & RHS are not equal
...so we cannot distribute an existential quantifier over a conjunction.