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BSPHCL Recruitment 2015 – Apply Online for 114 Administrator & Manager Posts Read more: BSPHCL Recruitment 2015 - Apply Online for 114 Administrator & Manager Posts http://www.freejobalert.com/bihar-state-power-holding-company-ltd/136140/#ixzz3bpQa9Xur

BSPHCL Recruitment 2015 – Apply Online for 114 Administrator & Manager Posts: Bihar State Power Holding Company Limited (BSPHCL) has released notification for the recruitment of 104 Chief Data Base Administrator, Data Base Administrator, IT Manager vacancies in Bihar State Power (Holding) Company Ltd. (BSPHCL), North Bihar Power Distribution Company Ltd. (NBPDCL), South Bihar Power Distribution Company Ltd. (SBPDCL), Bihar State Power Transmission Company Ltd. (BSPTCL) & Bihar State Power Generation Company Ltd. (BSPGCL). Eligible candidates may apply online from 29-05-2015 to 14-06-2015 by 05:00 P.M & send hard copy of application on or before 24-06-2015 by 05;00 P.M. Other details like age limit, educational qualification, selection process and how to apply are given below….
BSPHCL Vacancy Details:
Total No. of Posts: 114
Name of the Posts:
1. Chief Data Base Administrator: 05 Posts
2. Data Base Administrator: 08 Posts
3. IT Manager: 101 Posts
Age Limit: Candidates maximum age should be 45 years for post 1, 42 years for post 2, between 21-37 years for UR, 21-42 years for SC/ ST, 21-40 years for BC/ EBC/ Female (UR) as on 01-05-2015. Age relaxation is applicable 10 years for PH candidates as per rules.
Educational Qualification: Candidates should possess B.E/ B.Tech (Computer Science/ IT) with first class/ MCA from any Govt./ AICTE recognized Institution/ University with relevant experience.
Selection Process: Candidates will be selected based on work experience, written test & interview.
Application Fee: Candidates have to pay Rs. 1500/- (Rs. 375/- for SC/ ST of Bihar domicile) for post 1, 2, Rs. 1000/- (Rs. 250/- for SC/ ST of Bihar domicile) for post 3 deposited in the BSP(H)CL Current A/C No.- (Power Jyoti) 31963202219 in any branch of State Bank of India by filling a triplicate Challan. SC/ ST candidates of other States will be treated as UR candidates even for the purpose of Application fee.
How to Apply: Eligible candidates may apply online through the website www.bsphcl.bih.nic.in or www.applyfortest.com/bsphcl.aspx from 29-05-2015 to 14-06-2015 by 05:00 P.M & send hard copy of online application along with necessary certificates through ordinary post to- Post Box Number 12006; Cossipore Post Office Kolkata: 700002 on or before 24-06-2015 by 05:00 P.M. Superscribe the envelope with “name of the post & category”.
Instructions to Apply Online:
1. Log on through website www.bsphcl.bih.nic.in or http://www.applyfortest.com/bsphcl.aspx.
2. Click on “Online Application for the Post of CDBA, DBA and IT Manager with BSPHCL and its subsidiary company”.
3. Click on”Apply Online”.
4. Fill all the details carefully in two steps & submit the form.
5. Now take the printout of online application for future use.
Important Dates:
Starting Date to Apply Online: 29-05-2015.
Last Date to Apply Online: 14-06-2015 by 05:00 P.M.
Last Date for Submission of Hard copy of Application: 24-06-2015 by 05:00 P.M.
Downloading of Admit Card from the website: From 30-06-2015.
Tentative Date of Examination for Post 3: 05-07-2015 (1st sitting).
Tentative Date of Examination for Post 1 & 2: 05-07-2015 (2nd sitting).
For more details like category wise post reservation, remuneration and other information click on the link given below…

Truth Tables, Tautologies, and Logical Equivalence


Mathematics normally works with a two-valued logic: Every statement is either True or False. You can use truth tables to determine the truth or falsity of a complicated statement based on the truth or falsity of its simple components.
A statement in sentential logic is built from simple statements using the logical connectives $\lnot$ , $\land$ , $\lor$ , $\ifthen$ , and $\iff$ . I'll construct tables which show how the truth or falsity of a statement built with these connective depends on the truth or falsity of its components.
Here's the table for negation:
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & P & & $\lnot P$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & T & & F & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & F & & T & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$
This table is easy to understand. If P is true, its negation $\lnot P$ is false. If P is false, then $\lnot P$ is {\it true}.
$P \land Q$ should be true when both P and Q are true, and false otherwise:
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & P & & Q & & $P \land Q$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & T & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
$P \lor Q$ is true if either P is true or Q is true (or both). It's only false if both P and Q are false.
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & P & & Q & & $P \lor Q$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & F & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
Here's the table for logical implication:
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & P & & Q & & $P \ifthen Q$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
To understand why this table is the way it is, consider the following example:
"If you get an A, then I'll give you a dollar."
The statement will be true if I keep my promise and false if I don't.
Suppose it's true that you get an A and it's true that I give you a dollar. Since I kept my promise, the implication is {\it true}. This corresponds to the first line in the table.
Suppose it's true that you get an A but it's false that I give you a dollar. Since I didn't keep my promise, the implication is false. This corresponds to the second line in the table.
What if it's false that you get an A? Whether or not I give you a dollar, I haven't broken my promise. Thus, the implication can't be false, so (since this is a two-valued logic) it must be true. This explains the last two lines of the table.
$P \iff Q$ means that P and Q are equivalent. So the double implication is true if P and Q are both true or if P and Q are both false; otherwise, the double implication is false.
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & P & & Q & & $P \iff Q$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & T & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
You should remember --- or be able to construct --- the truth tables for the logical connectives. You'll use these tables to construct tables for more complicated sentences. It's easier to demonstrate what to do than to describe it in words, so you'll see the procedure worked out in the examples.
Remarks. 1. When you're constructing a truth table, you have to consider all possible assignments of True (T) and False (F) to the component statements. For example, suppose the component statements are P, Q, and R. Each of these statements can be either true or false, so there are $2^3 = 8$ possibilities.
When you're listing the possibilities, you should assign truth values to the component statements in a systematic way to avoid duplication or omission. The easiest approach is to use lexicographic ordering. Thus, for a compound statement with three components P, Q, and R, I would list the possibilities this way:
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & P & & Q & & R & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & T & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & F & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & T & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & T & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & F & & F & & F & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
2. There are different ways of setting up truth tables. You can, for instance, write the truth values "under" the logical connectives of the compound statement, gradually building up to the column for the "primary" connective.
I'll write things out the long way, by constructing columns for each "piece" of the compound statement and gradually building up to the compound statement.
Example. Construct a truth table for the formula $\lnot P \land (P \ifthen Q)$ .
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & $\lnot P$ & & $P \ifthen Q$ & & $\lnot P \land (P \ifthen Q)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & F & & T & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & F & & F & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }}\quad\halmos $$
A tautology is a formula which is "always true" --- that is, it is true for every assignment of truth values to its simple components. You can think of a tautology as a rule of logic.
The opposite of a tautology is a contradiction, a formula which is "always false". In other words, a contradiction is false for every assignment of truth values to its simple components.
Example. Show that $(P \ifthen Q) \lor (Q \ifthen P)$ is a tautology.
I construct the truth table for $(P \ifthen Q) \lor (Q \ifthen P)$ and show that the formula is always true.
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & $P \ifthen Q$ & & $Q \ifthen P$ & & $(P \ifthen Q) \lor (Q \ifthen P)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & T & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
The last column contains only T's. Therefore, the formula is a tautology.
Example. Construct a truth table for $(P \ifthen Q) \land (Q \ifthen R)$ .
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & R & & $P \ifthen Q$ & & $Q \ifthen R$ & & $(P \ifthen Q) \land (Q \ifthen R)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & F & & T & & F & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & T & & F & & T & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & F & & F & & T & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & F & & T & & F & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & F & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }}\quad\halmos $$
Example. Suppose
"$x > y$ " is true.
"$\displaystyle \int f(x)\,dx = g(x) + C$ " is false.
"Calvin Butterball has purple socks" is true.
Determine the truth value of the statement
$$(x > y \ifthen \int f(x)\,dx = g(x) + C) \ifthen\,\lnot (\hbox{Calvin Butterball has purple socks})$$
For simplicity, let
P = "$x > y$ ".
Q = "$\displaystyle \int f(x)\,dx = g(x) + C$ ".
R = "Calvin Butterball has purple socks".
I want to determine the truth value of $(P \ifthen Q) \ifthen\,\lnot R$ . Since I was given specific truth values for P, Q, and R, I set up a truth table with a single row using the given values for P, Q, and R:
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & R & & $P \ifthen Q$ & & $\lnot R$ & & $(P \ifthen Q) \ifthen\,\lnot R$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & T & & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
Therefore, the statement is true.
Two statements X and Y are logically equivalent if $X \iff Y$ is a tautology. Another way to say this is: For each assignment of truth values to the simple statements which make up X and Y, the statements X and Y have identical truth values.
From a practical point of view, you can replace a statement in a proof by any logically equivalent statement.
To test whether X and Y are logically equivalent, you could set up a truth table to test whether $X \iff Y$ is a tautology --- that is, whether $X \iff Y$ "has all T's in its column". However, it's easier to set up a table containing X and Y and then check whether the columns for X and for Y are the same.
Example. Show that $P \ifthen Q$ and $\lnot P \lor Q$ are logically equivalent.
$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & $P \ifthen Q$ & & $\lnot P$ & & $\lnot P \lor Q$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & T & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & F & & F & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$
Since the columns for $P \ifthen Q$ and $\lnot P \lor Q$ are identical, the two statements are logically equivalent. This tautology is called Conditional Disjunction. You can use this equivalence to replace a conditional by a disjunction.
There are an infinite number of tautologies and logical equivalences; I've listed a few below; a more extensive list is given at the end of this section.
$$\matrix{ \hbox{Double negation} & \lnot(\lnot P) \iff P \cr \hbox{DeMorgan's Law} & \lnot(P \lor Q) \iff (\lnot P \land \lnot Q) \cr \hbox{DeMorgan's Law} & \lnot(P \land Q) \iff (\lnot P \lor \lnot Q) \cr \hbox{Contrapositive} & (P \ifthen Q) \iff (\lnot Q \ifthen\,\lnot P) \cr \hbox{Modus ponens} & [P \land (P \ifthen Q)] \ifthen Q \cr \hbox{Modus tollens} & [\lnot Q \land (P \ifthen Q)] \ifthen\,\lnot P \cr}$$
When a tautology has the form of a biconditional, the two statements which make up the biconditional are logically equivalent. Hence, you can replace one side with the other without changing the logical meaning.
Example. Write down the negation of the following statements, simplifying so that only simple statements are negated.
(a) $(P \lor \lnot Q)$
$$\matrix{\lnot(P \lor \lnot Q) & \iff & \lnot P \land \lnot\lnot Q & \hbox{DeMorgan's law} \cr & \iff & \lnot P \land Q & \hbox{Double negation} \cr}\quad\halmos$$
(b) $(P \land Q) \ifthen R$
$$\matrix{\lnot[(P \land Q) \ifthen R] & \iff & \lnot[\lnot(P \land Q) \lor R] & \hbox{Conditional Disjunction} \cr & \iff & \lnot\lnot(P \land Q) \land \lnot R & \hbox{DeMorgan's law} \cr & \iff & (P \land Q) \land \lnot R & \hbox{Double negation} \cr}$$
I showed that $(A \ifthen B)$ and $(\lnot A \lor B)$ are logically equivalent in an earlier example.
Example. Use DeMorgan's Law to write the negation of the following statement, simplifying so that only simple statements are negated:
"Calvin is not home or Bonzo is at the movies."
Let C be the statement "Calvin is home" and let B be the statement "Bonzo is at the moves". The given statement is $\lnot C \lor B$ . I'm supposed to negate the statement, then simplify:
$$\matrix{\lnot(\lnot C \lor B) & \iff & \lnot\lnot C \land \lnot B & \hbox{DeMorgan's Law} \cr & \iff & C \land \lnot B & \hbox{Double negation} \cr}$$
The result is "Calvin is home and Bonzo is not at the movies".
Example. Use DeMorgan's Law to write the negation of the following statement, simplifying so that only simple statements are negated:
"If Phoebe buys a pizza, then Calvin buys popcorn."
Let P be the statement "Phoebe buys a pizza" and let C be the statement "Calvin buys popcorn". The given statement is $P \ifthen C$ . To simplify the negation, I'll use the Conditional Disjunction tautology which says
$$(P \ifthen Q) \iff (\lnot P \lor Q)$$
That is, I can replace $P \ifthen Q$ with $\lnot P \lor Q$ (or vice versa).
Here, then, is the negation and simplification:
$$\matrix{\lnot(P \ifthen C) & \iff & \lnot(\lnot P \lor C) & \hbox{Conditional Disjunction} \cr & \iff & \lnot\lnot P \land \lnot C & \hbox{DeMorgan's Law} \cr & \iff & P \land \lnot C & \hbox{Double negation} \cr}$$
The result is "Phoebe buys the pizza and Calvin doesn't buy popcorn".
Example. Replace the following statement with its contrapositive:
"If x and y are rational, then $x + y$ is rational."
By the contrapositive equivalence, this statement is the same as "If $x + y$ is not rational, then it is not the case that both x and y are rational".
Example. Show that the inverse and the converse of a conditional are logically equivalent.
Let $P \ifthen Q$ be the conditional. The inverse is $\lnot P \ifthen\,\lnot Q$ . The converse is $Q \ifthen P$ .
I could show that the inverse and converse are equivalent by constructing a truth table for $(\lnot P \ifthen\,\lnot Q) \iff (Q \ifthen P)$ . I'll use some known tautologies instead.
Start with $\lnot P \ifthen\,\lnot Q$ :
$$\matrix{\lnot P \ifthen\,\lnot Q & \iff & \lnot\lnot Q \ifthen \lnot\lnot P & \hbox{Contrapositive} \cr & \iff & Q \ifthen P & \hbox{Double negation} \cr}$$
Remember that I can replace a statement with one that is logically equivalent. For example, in the last step I replaced $\lnot\lnot Q$ with Q, because the two statements are equivalent by Double negation.
Example. Suppose x is a real number. Consider the statement
"If $x^2 = 4$ , then $x = 2$ ."
Construct the converse, the inverse, and the contrapositive. Determine the truth or falsity of the four statements --- the original statement, the converse, the inverse, and the contrapositive --- using your knowledge of algebra.
The converse is "If $x = 2$ , then $x^2 = 4$ ".
The inverse is "If $x^2 \ne 4$ , then $x \ne 2$ ".
The contrapositive is "If $x \ne 2$ , then $x^2 \ne 4$ ".
The original statement is false: $(-2)^2 = 4$ , but $-2 \ne 2$ . Since the original statement is eqiuivalent to the contrapositive, the contrapositive must be false as well.
The converse is true. The inverse is logically equivalent to the converse, so the inverse is true as well.

List of Tautologies

$$\matrix{ 1. \hfill & P \lor \lnot P \hfill & \hbox{Law of the excluded middle} \hfill \cr 2. \hfill & \lnot(P \land \lnot P) \hfill & \hbox{Contradiction} \hfill \cr 3. \hfill & [(P \ifthen Q) \land \lnot Q] \ifthen \lnot P \hfill & \hbox{ Modus tollens} \hfill \cr 4. \hfill & \lnot\lnot P \iff P \hfill & \hbox{Double negation} \hfill \cr 5. \hfill & [(P \ifthen Q) \land (Q \ifthen R)] \ifthen (P \ifthen R) \hfill & \hbox{Law of the syllogism} \hfill \cr 6. \hfill & (P \land Q) \ifthen P \hfill & \hbox{Decomposing a conjunction} \hfill \cr & (P \land Q) \ifthen Q \hfill & \hbox{Decomposing a conjunction} \hfill \cr 7. \hfill & P \ifthen (P \lor Q) \hfill & \hbox{Constructing a disjunction} \hfill \cr & Q \ifthen (P \lor Q) \hfill & \hbox{Constructing a disjunction} \hfill \cr 8. \hfill & (P \iff Q) \iff [(P \ifthen Q) \land (Q \ifthen P)] \hfill & \hbox{Definition of the biconditional} \hfill \cr 9. \hfill & (P \land Q) \iff (Q \land P) \hfill & \hbox{Commutative law for $\land$} \hfill \cr 10. \hfill & (P \lor Q) \iff (Q \lor P) \hfill & \hbox{Commutative law for $\lor$} \hfill \cr 11. \hfill & [(P \land Q) \land R] \iff [P \land (Q \land R)] \hfill & \hbox{Associative law for $\land$} \hfill \cr 12. \hfill & [(P \lor Q) \lor R] \iff [P \lor (Q \lor R)] \hfill & \hbox{Associative law for $\lor$} \hfill \cr 13. \hfill & \lnot(P \lor Q) \iff (\lnot P \land \lnot Q) \hfill & \hbox{DeMorgan's law} \hfill \cr 14. \hfill & \lnot(P \land Q) \iff (\lnot P \lor \lnot Q) \hfill & \hbox{DeMorgan's law} \hfill \cr 15. \hfill & [P \land (Q \lor R)] \iff [(P \land Q) \lor (P \land R)] \hfill & \hbox{Distributivity} \hfill \cr 16. \hfill & [P \lor (Q \land R)] \iff [(P \lor Q) \land (P \lor R)] \hfill & \hbox{Distributivity} \hfill \cr 17. \hfill & (P \ifthen Q) \iff (\lnot Q \ifthen \lnot P) \hfill & \hbox{Contrapositive} \hfill \cr 18. \hfill & (P \ifthen Q) \iff (\lnot P \lor Q) \hfill & \hbox{Conditional disjunction} \hfill \cr 19. \hfill & [(P \lor Q) \land \lnot P] \ifthen Q \hfill & \hbox{Disjunctive syllogism} \hfill \cr 20. \hfill & (P \lor P) \iff P \hfill & \hbox{Simplification} \hfill \cr 21. \hfill & (P \land P) \iff P \hfill & \hbox{Simplification} \hfill \cr}$$
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