Week 1 Math230

A) A) x A) A) x

B) ¬∀x

C)

D) Some people did not take medication or placebo.

) A) →

B) -∀x∧

C) n

D) It is true some people did not take placebo or did not have migraines

C) A) ∧

B)

C)

D) It is true all patients did not have migraines or did not take placebo.

( can you please ignore the → above the x varibles they are their by mistake the current Ide enviorment im in is new to me, i didnt do it again haha. )

)

I will show the equivalence by verifying if each function on the sides of the = sign are equivalent to the other side of the = sign.

A) ok so i have one side as
we can distribute the - sign into our function for x for P and Q
we have ∨ - - )
we can use our double negation rule for Q and turn it positive.
thus we have an equivalence to the other side.
our and symbol is flipped as we have done de morgans law.
∨ )
these are equivalent expressions!

B) ok so we have one side as -∀x
We can begin by using de morgans law to see if these are recipicoral meaning they are equivalent when we inverse the properties i learned this in pre calculus. Which is why we use de morgans law.
we distribute the - sign from ∀ into our function for P and Q
we have (-- the → implies an and so we can use ∧ and symbol negative sign carries to the Q as well giving us
we use double negative to give us a positive .
after solving with de morgans law we do not have
logical statements that are inverses of each other, fundamentally
these functions are not equivalent to each other!

C)
Is logically equivalent. I solved this mentally but I can share how to solve this.
ok so we use De morgans law to get the inverse and with this law we flip our expression to be the opposite of what they are.

For -∀ we distubute the negative sign into the pararntheis and we flip the ∀ to an ∃
we have almost the same expression on the other side
and when we use our double negative law they are the same however
we now have to flip our symbols as that is fundemental to de morgans law.
These are logically equivalent !

) A) is true i read the function as all instances of x and y where the y and x are not = to each other to be True. This is known to be correct by looking at the truth table for sections where these all say F so this is true. as the case no one sent mail to themselves.

B) Is true as the predicate is M which is saying the emails did not happen to be sent to all values of x and some for y. this is true , there is instances where mail is not sent for all vaules of x and some for y. to see this we can go horizontally across the x vaules truth tables our symbol ∀ suggests its for all and the - m implies no mail was given, this is read the row has no x vaules where all mail was given in truth table we have T and Fs so this cant be true implying our expression is correct. for our y it states for some, there is no mail which is true, we go down from the y axis on the truth table and see there is T and Fs for each column.

C) is not true
the predicate M suggest the x and y values got mail.
our expressions suggests some vaules of x got mail and all vaules of y got mail.
while its true some x s got mail, all the ys did not get mail.
this quantified statement is not true!