Answer
A) ∃xV(x) , ∃x( S(x) ∧ V(x)) , ∃x( S(x) ∧ V(x,Vietnam))
B) ∃xH(x) , ∃x( S(x) → ~H(x)) , ∃x( S(x) → ~H(x,Hindi))
C) ∃x( J(x) ∧P(x) ∧ C(x)), ∃x( S(x) → ((J(x) ∧P(x) ∧ C(x))) , ∃x(S(x) → (K(x,Java) ∧ K(x,Prolog) ∧ K(x,C++))).
D) ∀xT(x), ∀x( S(x) ∧ T(x)), ∀x( S(x) ∧ T(x,Thai)).
E) ∃xH(x) , ∃x( S(x) ∧ ~H(x)) , ∃x( S(x) ∧ ~H(x,hockey))
Work Step by Step
for all of the answers, we will take S(x) to be "x is in your school
A) if we let V(x)b be "x has lived in Vietnam" then we have ∃xV(x) if the domain is just your schoolmates.
Or ∃x( S(x) ∧ V(x)) if the domain is all people.
If we let V(x,y) to be "x has lived in place y" then we can write the expression as ∃x( S(x) ∧ V(x,Vietnam)).
B) if we let H(x) be "x can speak Hindi" then ∃x~H(x) if the domain is just your schoolmates.
Or ∃x( S(x) → ~H(x)) if the domain is all people.
If we let H(x,y) to be "x can speak the language y" then we can write the expression as ∃x( S(x) → ~H(x,Hindi)).
C) if we let J(x) be "x knows Java" and P(x) be "x knows Prolog" and C be "x knows C++" then ∃x( J(x) ∧P(x) ∧ C(x)) if the domain is just your schoolmates.
Or ∃x( S(x) → ((J(x) ∧P(x) ∧ C(x))) if the domain is all people.
If we let K(x,y) to be "x knows the programming language y" then we can write the expression as ∃x( S(x) → (K(x,Java) ∧ K(x,Prolog) ∧ K(x,C++))).
D) B) if we let T(x) be "x enjoys Thai food" then ∀xT(x) if the domain is just your schoolmates.
Or ∀x( S(x) ∧ T(x)) if the domain is all people.
If we let T(x,y) to be "x enyojs the food y" then we can write the expression as ∀x( S(x) ∧ T(x,Thai)).
E) if we let H(x)b be "x plays hockey" then we have ∃x~H(x) if the domain is just your schoolmates.
Or ∃x( S(x) ∧ ~H(x)) if the domain is all people.
If we let H(x,y) to be "x plays the game y" then we can write the expression as ∃x( S(x) ∧ ~H(x,hockey)).
