Imagine someone comes up to you and claims: "Someone in the family makes bread". How do you disprove this claim?
Well the only way to do so is to prove that EVERY person in that given family does NOT make bread. If theres AT LEAST one person, then your efforts are in vain and you cannot disprove it. So that's why you need to check EVERYONE.
So this means that the negation of "Someone in the family makes bread" is
Everyone in the family does NOT make bread
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# 2
p: Mary is on the bus.
q: April is in the car.
r: Stan is at the zoo
Since r: Stan is at the zoo, this means that
~r: Stan is NOT at the zoo
Also, because q: April is in the car, this tells us that
~q: April is NOT in the car
Now, translate p v ~r:
p v ~r: Mary is on the bus OR Stan is NOT at the zoo
Now connect this with ~q in the form of IF...THEN (since --> means IF..THEN) to get:
IF Mary is on the bus OR Stan is NOT at the zoo, THEN April is NOT in the car.
So the compound statement (p v ~ r) --> ~q translates to:
(p v ~ r) --> ~q : IF Mary is on the bus OR Stan is NOT at the zoo, THEN April is NOT in the car.
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# 3
Let
p: Today is Friday.
q: Tomorrow is not the day to go shopping.
Since
p: Today is Friday.
this means that
~p: Today is NOT Friday.
Now recall that arguments of the form
p -> q
are read as "IF p, THEN q"
In the case of "IF tomorrow is not the day to go shopping, THEN today is not Friday", we can see that the statement q is the antecendent and ~p is the consequent.
So the given statement translates to: q -> ~p
So
q -> ~p: "IF tomorrow is not the day to go shopping, THEN today is NOT Friday"
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# 4
First, make a table for p ^ q. Remember, p ^ q means "p AND q". So the compound statement "p ^ q" is ONLY true if BOTH p and q are true. Otherwise, "p ^ q" is false.
| p | q | p^q |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
Now to form ~(p ^ q) (which reads "NOT p and q"). So if p ^ q is true, then ~(p ^ q) is false. Similarly, if p ^ q is false, then ~(p ^ q) is true. In other words, just change the make the truth values the opposite of "p ^ q" like so:
| p | q | p^q | ~(p^q) |
| T | T | T | F |
| T | F | F | T |
| F | T | F | T |
| F | F | F | T |
Notice how the 3rd and 4th columns are opposites of one another.
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# 5
The statement "IF the sun is shining, THEN it will NOT rain" is in the form "If p, then NOT q" where
p: the sun is shining
q: it will rain
So the symbolic form of the conditional statement is: p -> ~q
Converse: To find the converse of p -> q, simply switch the antecendent and consequent to get q -> p
In this case, swap p and ~q to get ~q -> p
So the converse of "IF the sun is shining, THEN it will NOT rain" is
"IF it will NOT rain, THEN the sun is shining"
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Inverse: To form the inverse of p -> q, negate every individual statement to get: ~p -> ~q
For this problem, negate p to get ~p and negate ~q to get q. So the symbolic inverse is ~p -> q
So the inverse of "IF the sun is shining, THEN it will NOT rain" is
"IF the sun is NOT shining, THEN it will rain"
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Contrapositive: To form the contrapositive of p -> q, switch p and q AND negate both statements to get ~q -> ~p
For this problem, negate p to get ~p and negate ~q to get q. Now swap the two to get: q -> ~p
So the contrapositive of "IF the sun is shining, THEN it will NOT rain" is
"IF it will rain, THEN the sun is NOT shining"
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# 6
Let
p: a tree has green leaves
q: a tree produces oxygen
So the argument is
p ^ q
p
------
:. q
Let's make a truth table
Note: the premises are in red and the conclusion is in blue
| p | q | p ^ q | p | q |
| T | T | T | T | T |
| T | F | F | T | F |
| F | T | F | F | T |
| F | F | F | F | F |
Since there are no rows that have all true premises and a false conclusion, this means that we cannot prove that the argument is invalid. So the argument is valid.
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# 7
First, draw a large circle to denote the set of all people who posses golf carts
Inside this large circle, draw two smaller circles (that are completely inside this larger circle). Make sure that they overlap. Why? Well there is a possibility that SOME golfers are members of club A (or vice versa). However, we cannot be 100% certain that all golfers are members (or vice versa) and we cannot be certain that no golfer is a member of club A. So this covers all of the possible options.
Note: if we knew for certain that all members of club A are golfers, then we would draw the set of golfers inside the set of club A members.
So the drawing looks like this:
Since we CANNOT be 100% whether all golfers are members of club A, this means that the following syllogism is invalid. In other words, we cannot logically deduce the conclusion given the following premises.
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# 8
By convention since " p is false, q is true, and r is true", we denote these variables by
p = F,
q = T, and
r = T
~q <--> [~ r v (p ^ q)] ... Start with the given logical compound statement.
~T <--> [~ T v (F ^ T)] ... Plug in p = F, q = T, and r = T
F <--> [F v (F ^ T)] ... Evaluate ~T to get F (~T means NOT true or false)
F <--> [F v F] ... Evaluate F ^ T to get F (F ^ T means false AND true which is false)
F <--> F ... Evaluate F v F to get F (F v T means false OR false which is false)
T ... Evaluate F <--> F to get T
Note: p <--> q is only true if the truth values of p and q are equal. In other words, p <--> q is true if p and q are true (or if p and q are false). Think of the connective "<-->" as the "equal sign" (technically it's not, but it's a good analogy)
So ~T <--> [~ T v (F ^ T)] = T
This means that ~q <--> [~ r v (p ^ q)] is true when "p is false, q is true, and r is true"
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# 9
Let
p: Rembrandt was a famous painter
q: all prime numbers are odd
The statement "Rembrandt was a famous painter and all prime numbers are odd" translates to
p ^ q
Since Rembrandt was indeed a famous painter, this means that 'p' is true. However, since not all prime numbers are odd (2 is even AND is prime), this means that 'q' is false. So we know that p = T and q = F
So "p ^ q" then becomes "T ^ F" which evaluates to false (since BOTH statements must be true for p ^ q to be true).
So the statement "Rembrandt was a famous painter and all prime numbers are odd" is false.
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# 10
Remember, one of DeMorgan's Laws is
~(A ^ B) = ~A v ~B
In this case, the expression ~(p ^ q) expands to
~(p ^ q) = ~p v ~q
From inspection (ie just looking at it), we can easily see that the two are NOT equal. However let's make a truth table comparing ~p v ~q and ~p ^ ~q
Truth table for ~p v ~q:
| p | q | ~p | ~q | ~p v ~q |
| T | T | F | F | F |
| T | F | F | T | T |
| F | T | T | F | T |
| F | F | T | T | T |
Truth table for ~p ^ ~q:
| p | q | ~p | ~q | ~p ^ ~q |
| T | T | F | F | F |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |
Clearly, the last columns from both tables are NOT identical. So this shows that ~p v ~q is NOT equal to ~p ^ ~q
Consequently, ~(p ^ q) is NOT equal to ~p ^ ~q
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# 11
Let
p: Today is Monday
q: Tomorrow is Tuesday
So this means that
~p: Today is NOT Monday
~q: Tomorrow is NOT Tuesday
a) The statement "If today is Monday, then tomorrow is Tuesday" translates to:
p --> q
Since "-->" means "IF...THEN"
Now let's make a truth table for p --> q:
| p | q | p --> q |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Note: p --> q is only false when p is true and q is false
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b) The statement "If today is not Monday, then tomorrow is not Tuesday" translates to:
~p --> ~q
Since it is a conditional statement
Now let's make a truth table for ~p --> ~q:
| p | q | ~p | ~q | ~p --> ~q |
| T | T | F | F | T |
| T | F | F | T | T |
| F | T | T | F | F |
| F | F | T | T | T |
Since the last columns of part a) and part b) are NOT identical, this means that p --> q does NOT equal ~p --> ~q
So statements a) and b) are NOT equivalent
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c) The statement "If tomorrow is not Tuesday, then today is not Monday" translates to:
~q --> ~p
Let's make a truth table for ~q --> ~p:
| p | q | ~p | ~q | ~q --> ~p |
| T | T | F | F | T |
| T | F | F | T | F |
| F | T | T | F | T |
| F | F | T | T | T |
Take note how the last column of this part is identical to the last column of part a). So this shows that statements a) and c) are equivalent.