Q.No.1
Assume that for the truth values
P=F, q=T, r=F
Show that the proposition
~(pvq)^~(~pvr) is false.
~(pvq)^~(~pvr)
F FTT F F TFTF
false
i want to solve the following questions:
Q.No.1
Assume that for the truth values
P=F, q=T, r=F
Show that the proposition
~(pvq)^~(~pvr) is false.
Q.No.2
Formulate the arguments symbolically and test its validity using truth table.
If you invest in the auto industry, then you get rich.
You didn't invest in the auto industry.
Therefore, you didn't get rich.
Where p = "You invest in the auto industry."
q = "You get rich."
Q.No.3
Illustrate the distributive law AÇ(BÈC)=(AÇB)È(AÇC) with Venn diagrams
pleae help me in solving these questions.
Azhar
Have you had a look around at wikipedia? (Eg. Logical connective - Wikipedia, the free encyclopedia) But there are many other entries which deal with this topic, such as 'propositional logic', 'boolean algebra.' and even 'truth table.'
So, the first proposition is 'if p then q' : ( )Formulate the arguments symbolically and test its validity using truth table.
If you invest in the auto industry, then you get rich.
You didn't invest in the auto industry.
Therefore, you didn't get rich.
Where p = "You invest in the auto industry."
q = "You get rich."
Then have a look at the corresponding truth table (implication). The other two propositions are similar to the first proposition. Remember that not p (which in the truth table is the equivalent of a truth 'value' of F(alse).)
If you look find out what a Venn diagram is, you will be able to do the third part. Venn diagrams are very good.
Q.No.2
Formulate the arguments symbolically and test its validity using truth table.
If you invest in the auto industry, then you get rich.
You didn't invest in the auto industry.
Therefore, you didn't get rich.
Where p = "You invest in the auto industry."
q = "You get rich."
p->q
~p
therefore ~q
[(p->q)->~p]->~q
T T FT FT
T F FT TF
F T TF FT
F F TF TF
[(p->q)->~p]->~q
T T T FT FT
T F F FT TF
F T T TF FT
F T F TF TF
[(p->q)->~p]->~q
T T T F FT FT
T F F T FT TF
F T T T TF FT
F T F T TF TF
[(p->q)->~p]->~q
T T T F FT T FT
T F F T FT T TF
F T T T TF F FT
F T F T TF T TF
SO the answer is False
Question 2
Definition: A proposition is a statement that is either true or false:
There are 2 propositions:
1. P:= you invest in the auto industry
2. Q:= you [become] rich
So, each proposition can be either T(rue) or F(alse). Ie. If you did invest in the auto industry the P is True, if not, then False.
So, there are different ways to combine two propositions, for example:
P AND Q,
P OR Q, etc.
Each operator has its own truth table.
You're interested in P IMPLIES Q, which is often presented as "if P then Q"
The truth table for is:
P | Q |
-------------------
T | T | T
T | F | F
F | T | T
F | F | T
So, now you could say, NOT P Q is T(rue) etc. Which would mean 1. (did not invest in the auto industry) 2. (becomes rich) True