GATE Overflow - Recent questions and answers in Mathematical Logic
http://gateoverflow.in/qa/mathematics/discrete-mathematics/mathematical-logic
Powered by Question2AnswerAnswered: Gate Question for CSE 1995 In Propositional logic
http://gateoverflow.in/135004/gate-question-for-cse-1995-in-propositional-logic?show=135005#a135005
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=8575549387575021324"></p>
<p>Answer is option <strong>D</strong></p>Mathematical Logichttp://gateoverflow.in/135004/gate-question-for-cse-1995-in-propositional-logic?show=135005#a135005Tue, 27 Jun 2017 20:26:59 +0000Answered: Propositional Logic
http://gateoverflow.in/134953/propositional-logic?show=134972#a134972
proposition is a declarative sentence that is either true or false but can not be both<br />
<br />
here x is a variable representing a number<br />
<br />
if we put x=5 then it is true and if we put any other value excluding x=5 then it will be false<br />
<br />
ie. unless a specific value is given to x we can not say whether it is true or false, <br />
<br />
so it is not proportionMathematical Logichttp://gateoverflow.in/134953/propositional-logic?show=134972#a134972Tue, 27 Jun 2017 15:26:39 +0000generating function
http://gateoverflow.in/134888/generating-function
how many integer solution to the equation a+b+c=6 satisfy -1<=a<=2 and 1<=b,c<=4?Mathematical Logichttp://gateoverflow.in/134888/generating-functionTue, 27 Jun 2017 05:16:43 +0000Answered: Rosen
http://gateoverflow.in/133952/rosen?show=134833#a134833
<p>P=you can ride the roller coaster
<br>
<br>
Q=you are under 4 feet tall.
<br>
<br>
R=You are older than 16 years old
<br>
<br>
use of if : <strong>A if B</strong> $\equiv B \rightarrow A$
<br>
<br>
use of unless : <strong>A unless B</strong> $\equiv \bar{B} \rightarrow A \equiv ~~B \vee A$
<br>
<br>
now,
<br>
<br>
$\bar{Q} \rightarrow \bar{P}$ <strong>unless </strong>$R$
<br>
<br>
$\equiv Q \vee \bar{P} \vee R$
<br>
<br>
$\equiv Q \vee R \vee \bar{P}$
<br>
<br>
$\equiv$ <strong>not</strong> $(Q \wedge \bar{R}) \vee \bar{P}$
<br>
<br>
$\equiv (Q \wedge \bar{R}) \rightarrow \bar{P}$</p>Mathematical Logichttp://gateoverflow.in/133952/rosen?show=134833#a134833Mon, 26 Jun 2017 13:01:35 +0000Gateoverflow pdf
http://gateoverflow.in/134677/gateoverflow-pdf
<p><img alt="" height="500" src="http://gateoverflow.in/?qa=blob&qa_blobid=12818629108791117556" width="500"></p>Mathematical Logichttp://gateoverflow.in/134677/gateoverflow-pdfSun, 25 Jun 2017 07:51:53 +0000Answered: Probability
http://gateoverflow.in/134654/probability?show=134658#a134658
$\begin{align*} &\Rightarrow P\left ( A' \cup B' \right ) = {\color{red}{P\left ( A' \right ) + P\left ( B' \right )}} - P\left ( A' \cap B' \right ) \\ &\Rightarrow P\left [ \left ( A \cap B \right )' \right ] = {\color{red}{P\left ( A' \right ) + P\left ( B' \right )}} - P \left [ \left ( A \cup B \right )' \right ] \\ &\Rightarrow 0.75 = {\color{red}{P\left ( A' \right ) + P\left ( B' \right )}} - 0.4 \\ &\Rightarrow {\color{red}{P\left ( A' \right ) + P\left ( B' \right )}} = 1.15 \end{align*}$Mathematical Logichttp://gateoverflow.in/134654/probability?show=134658#a134658Sun, 25 Jun 2017 01:59:07 +0000Answered: GATE_2015_ECE Matrices
http://gateoverflow.in/118139/gate_2015_ece-matrices?show=134364#a134364
<p style="text-align:center"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=15431600360778197932"></p>
<p>Answer</p>Mathematical Logichttp://gateoverflow.in/118139/gate_2015_ece-matrices?show=134364#a134364Thu, 22 Jun 2017 16:53:50 +0000Answered: [Discrete Maths] Graph Theory Rosen,Chromatic number
http://gateoverflow.in/132851/discrete-maths-graph-theory-rosen-chromatic-number?show=134190#a134190
3 is the correct answer for bothMathematical Logichttp://gateoverflow.in/132851/discrete-maths-graph-theory-rosen-chromatic-number?show=134190#a134190Wed, 21 Jun 2017 13:56:46 +0000Answered: [Discrete maths] Exponent generating function
http://gateoverflow.in/133771/discrete-maths-exponent-generating-function?show=134116#a134116
<p><span class="marker">solution for option d:</span></p>
<p><span class="marker"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11271631214124629879"></span></p>
<p><span class="marker">solution for option e:</span></p>
<p><span class="marker"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11781703958118828656"></span></p>Mathematical Logichttp://gateoverflow.in/133771/discrete-maths-exponent-generating-function?show=134116#a134116Wed, 21 Jun 2017 05:48:24 +0000Answered: [Discrete Maths] Counting
http://gateoverflow.in/132473/discrete-maths-counting?show=134115#a134115
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=9177115529628536683"></p>
<p> </p>
<p>please tell me whats wrong with my approach</p>Mathematical Logichttp://gateoverflow.in/132473/discrete-maths-counting?show=134115#a134115Wed, 21 Jun 2017 05:28:07 +0000Answered: TIFR2011-A-1
http://gateoverflow.in/237/tifr2011-a-1?show=133968#a133968
<p>let the variable we use are
<br>
W : wages are raised
<br>
P : prices are raised
<br>
I : there is inflation
<br>
G : government must regulate it
<br>
S : people will suffer
<br>
U : government will be unpopular</p>
<p>"If either wages or prices are raised, there will be inflation" : (W+P)->I ...(1)
<br>
"If there is inflation, then either the government must regulate it or the people will suffer" : I->(G+S) ...(2)
<br>
"If the people suffer, the government will be unpopular" : S->U ...(3)
<br>
"Government will not be unpopular" : U' ...(4)</p>
<p>options are :
<br>
A. "People will not suffer" : S'
<br>
B. "If the inflation is not regulated, then wages are not raised" : I'->W'
<br>
C. "Prices are not raised" : P'
<br>
D. "If the inflation is not regulated, then the prices are not raised" : I'->P'
<br>
E. "Wages are not raised" : W'</p>
<p>using (3) and (4) we get
<br>
S' ...(5)
<br>
which is option A <strong>so option A is Valid</strong>
<br>
from (1), we do contrapositive of it
<br>
I'->(W+P)' = I'->(W'.P')= (I'->W').(I'->P') which is options B and D respectively <strong>so options B and D are also valid </strong></p>
<p><strong>Hence options A, </strong><strong>B</strong><strong> and D are valid </strong>
<br>
</p>Mathematical Logichttp://gateoverflow.in/237/tifr2011-a-1?show=133968#a133968Tue, 20 Jun 2017 06:21:00 +0000Answered: TIFR2012-A-2
http://gateoverflow.in/20939/tifr2012-a-2?show=133960#a133960
"((If Mr.M is guilty, then no witness is lying) unless he is afraid)" : <br />
=G->¬∃x(W(x)∧L(x)) unless ∃x(A(x))<br />
=¬∃x(A(x))->G->¬∃x(W(x)∧L(x))<br />
=(¬∃x(A(x))∧G)->¬∃x(W(x)∧L(x))<br />
=¬∃x((A(x))∧G)->(W(x)∧L(x)))<br />
=¬∃x((A(x))∧G)->W(x)) ∧ ¬∃x((A(x))∧G)->L(x))<br />
<br />
"There is a witness who is afraid" :<br />
=∃x(W(x)∧A(x))<br />
=∃xW(x) ∧ ∃xA(x)<br />
<br />
¬∃x(((A(x))∧G)->W(x)) ...(1)<br />
¬∃x(((A(x))∧G)->L(x)) ...(2)<br />
∃xW(x) ...(3)<br />
∃xA(x) ...(4)<br />
<br />
using (1) and (3) we get<br />
¬∃x((A(x))∧G)) ...(5) <br />
<br />
No further simplication happen <br />
Hence Ans is CMathematical Logichttp://gateoverflow.in/20939/tifr2012-a-2?show=133960#a133960Tue, 20 Jun 2017 05:08:41 +0000Answered: TIFR2010-A-4
http://gateoverflow.in/18212/tifr2010-a-4?show=133954#a133954
Let us denote sentences with variables <br />
F : Bank receipt is forged <br />
L : Mr M is liable <br />
B : He will go bankrupt <br />
M : Bank loan him money <br />
<br />
"If the bank receipt is forged, then Mr. M is liable" : F->L ..(1)<br />
"If Mr. M is liable, he will go bankrupt" : L->B ...(2)<br />
"If the bank will loan him money, he will not go bankrupt" : M->B' ...(3)<br />
"The bank will loan him money" : M ...(4) <br />
<br />
optuions are :<br />
a. "Mr. M is liable" : L<br />
b. "The receipt is not forged" : F'<br />
c. "Mr. M will go bankrupt" : B<br />
d. "The bank will go bankrupt" : NOT GETTING THIS <br />
e. None of the above<br />
<br />
From (3) and (4) , using modus ponens we get <br />
B' ...(5) <br />
From (2) and (5) using modus tollens we get <br />
L' ...(6)<br />
From (1) and (6) using modus tollens we get <br />
F' which is equivalent to option B.<br />
<br />
Hence correct ans is BMathematical Logichttp://gateoverflow.in/18212/tifr2010-a-4?show=133954#a133954Tue, 20 Jun 2017 03:40:59 +0000Diameter of a graph and tree
http://gateoverflow.in/133949/diameter-of-a-graph-and-tree
Why there is a difference between diameter of a graph and tree?<br />
<br />
Diameter of a tree as i have read is the maximum path between two vertices(number of edges between two vertices)<br />
<br />
But for tree it says number of nodes on the longest path.<br />
<br />
But tree is a graph so why cant i find the diameter of tree in similar way?Mathematical Logichttp://gateoverflow.in/133949/diameter-of-a-graph-and-treeTue, 20 Jun 2017 01:43:39 +0000Answered: [Discrete Maths] Group Theory
http://gateoverflow.in/133264/discrete-maths-group-theory?show=133921#a133921
Here as it is intersection so the order mmsut divide both H and K<br />
<br />
here 3,20,4 does not divide 10<br />
<br />
B,C,E should be the answerMathematical Logichttp://gateoverflow.in/133264/discrete-maths-group-theory?show=133921#a133921Mon, 19 Jun 2017 17:34:26 +0000Answered: [Discrete Maths] Function,Rosen p-154.1.c
http://gateoverflow.in/132557/discrete-maths-function-rosen-p-154-1-c?show=133904#a133904
$R \to R$ is mentioned in question meaning the domain and range of $f$ are the set of all real numbers. Now, $f(-2) = \sqrt{-2}$ is not a real number and thus $f$ is no longer a valid function.Mathematical Logichttp://gateoverflow.in/132557/discrete-maths-function-rosen-p-154-1-c?show=133904#a133904Mon, 19 Jun 2017 15:33:02 +0000Answered: [Discrete Maths] predicate Logic
http://gateoverflow.in/133755/discrete-maths-predicate-logic?show=133878#a133878
<pre>
<strong><code>"Whenever there is an active alert, all queued messages are transmitted."</code></strong></pre>
<p>Whenever -> Condition Statement.</p>
<p>There is... -> Existential condition.</p>
<p>All -> Universal condition.</p>
<p>So it becomes:</p>
<p>"If there is an active alert, all queued messages are transmitted" -> If <strong>there exists</strong> an <strong>alert which is active</strong>, then if <strong>for every</strong> message which is queued then the message is transmitted."</p>
<p>I translated the conditional and quantifiers perfectly, there may be some mistake in English but that's not important.</p>Mathematical Logichttp://gateoverflow.in/133755/discrete-maths-predicate-logic?show=133878#a133878Mon, 19 Jun 2017 13:19:27 +0000Answered: GATE2000-2.7
http://gateoverflow.in/654/gate2000-2-7?show=133867#a133867
a ⇔ ( b V- b) = and a ⇔ True means both a and True are equivalent <br />
b ⇔c means both b and c are equivalent <br />
<br />
(a ∧ b) → (a ∧ c) ∨ d<br />
=(True ∧ b) → (True ∧ c) ∨ d (a ⇔ True)<br />
= b → c ∨ d<br />
=- b ∨ c ∨ d<br />
=- b ∨ b ∨ d (b ⇔c)<br />
= True ∨ d <br />
=True<br />
<br />
Hence ans is AMathematical Logichttp://gateoverflow.in/654/gate2000-2-7?show=133867#a133867Mon, 19 Jun 2017 12:33:16 +0000Answered: GATE1991_03,xii
http://gateoverflow.in/526/gate1991_03-xii?show=133857#a133857
" F1∧F2→F3 and F1∧F2→∼F3 are both tautologies " it is possible in 2 cases<br />
case 1) True→True<br />
case 2) False→False/True<br />
here F3 is in both F3 and ∼F3 form so only case 2) will apply<br />
so F1∧F2 is False means F1=False and F2=False<br />
<br />
(a). "Both F1 and F2 are tautologies" is INCORRECT<br />
<br />
(b). "The conjunction F1∧F2 is not satisfiable" is INCORRECT becoz for being satisfiable atleast one possibility of F1 and F2 should be True<br />
<br />
(c). "Neither is tautologous" is INCORRECT as both are Tautologies<br />
<br />
(d). "Neither is satisfiable " is INCOORECT as both are Tautologies so also satisfiable<br />
<br />
Hence correct ans is BMathematical Logichttp://gateoverflow.in/526/gate1991_03-xii?show=133857#a133857Mon, 19 Jun 2017 11:46:04 +0000Answered: [Discrete Maths] Dual of a function
http://gateoverflow.in/133131/discrete-maths-dual-of-a-function?show=133850#a133850
<p>finding dual means shifting from positive logic to negative logic and vice versa . hence 0 will change to 1 , 1 to 0 , and to or .</p>
<p>consider the table for a<=b and a^b=a</p>
<p>e.g</p>
<table border="1" cellpadding="1" cellspacing="1" style="height:133px; width:506px">
<tbody>
<tr>
<td>a</td>
<td>b</td>
<td>a^b</td>
</tr>
<tr>
<td>0</td>
<td>0</td>
<td>0</td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>not valid because a>b</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
</tr>
</tbody>
</table>
<p>now consider the dual of above table</p>
<table border="1" cellpadding="1" cellspacing="1" style="width:500px">
<tbody>
<tr>
<td>a</td>
<td>b</td>
<td>a or b =a</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>1</td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td>not valid</td>
</tr>
<tr>
<td>0</td>
<td>0</td>
<td>0</td>
</tr>
</tbody>
</table>
<p>we can clearly see that if a or b =a then a>= b</p>
<p>therefore option d is correct</p>Mathematical Logichttp://gateoverflow.in/133131/discrete-maths-dual-of-a-function?show=133850#a133850Mon, 19 Jun 2017 11:07:29 +0000[Discrete Maths] Generating functions
http://gateoverflow.in/133773/discrete-maths-generating-functions
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=11849429701953093316"></p>Mathematical Logichttp://gateoverflow.in/133773/discrete-maths-generating-functionsMon, 19 Jun 2017 02:10:59 +0000Answered: [Discrete maths] permutations
http://gateoverflow.in/132560/discrete-maths-permutations?show=133751#a133751
For first position every 12 horses have equal possibility=12<br />
<br />
Since 1st position is acquired by one of the 12 horses,so other 11 horses have equal possibility for second position=11<br />
<br />
Since 2nd position is acquired by one of the 11 horses,so other 10 horses have equal possibility for second position=10<br />
<br />
so,12*11*10 would be total possibilities.Mathematical Logichttp://gateoverflow.in/132560/discrete-maths-permutations?show=133751#a133751Sun, 18 Jun 2017 17:02:24 +0000Answered: GATE2014-1-53
http://gateoverflow.in/1933/gate2014-1-53?show=133654#a133654
<p>propositional logic formulas is TRUE when exactly two of p,q and r are TRUE means possibilities are pq or qr or pr to be TRUE
<br>
<strong>option A)</strong>
<br>
((p↔q)∧r)∨(p∧q∧∼r)
<br>
=((T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
<br>
=((T)∧r)∨(T∧∼r)
<br>
=(r)∨(∼r)
<br>
=T
<br>
It doesn't depend upon the value of r, so, if the value of r is T then it also to be T so the condition exactly 2 are T is false. Hence <strong>option A is FALSE.</strong></p>
<p><strong>option B)</strong>
<br>
(∼(p↔q)∧r)∨(p∧q∧∼r)
<br>
=(∼(T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
<br>
=(∼(T)∧r)∨(T∧∼r)
<br>
=(F∧r)∨(T∧∼r)
<br>
=(F)∨(∼r)
<br>
=∼r
<br>
so if r=F then this is T so condition exactly 2 are T is true</p>
<p> (∼(p↔q)∧r)∨(p∧q∧∼r)
<br>
=(∼(T↔q)∧T)∨(T∧q∧∼T) ( take p=T and r=T)
<br>
=(∼(q)∧T)∨(F)
<br>
=∼(q)∨(F)
<br>
=∼q
<br>
so if q=F then this is T so condition exactly 2 are T is true</p>
<p>(∼(p↔q)∧r)∨(p∧q∧∼r)
<br>
=(∼(p↔T)∧T)∨(p∧T∧∼T) ( take q=T and r=T)
<br>
=(∼(p)∧T)∨(F)
<br>
=∼(p)∨(F)
<br>
=∼p
<br>
so if p=F then this is T so condition exactly 2 are T is true</p>
<p><strong>Hence option B is CORRECT.</strong></p>
<p><strong>option C)</strong>
<br>
((p→q)∧r)∨(p∧q∧∼r)
<br>
=((T→T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
<br>
=((T)∧r)∨(T∧∼r)
<br>
=(r)∨(∼r)
<br>
=T
<br>
It doesnt depend upon value of r , so , if value of r is T then it also to be T so the condition exactly 2 are T is false . <strong>Hence option C is FALSE.</strong></p>
<p><strong>option D)</strong>
<br>
(∼(p↔q)∧r)∧(p∧q∧∼r)
<br>
=(∼(T↔T)∧r)∧(T∧T∧∼r) ( take q=T and r=T)
<br>
=(∼(T)∧r)∧(T∧∼r)
<br>
=(F∧r)∧(T∧∼r)
<br>
=F∧(∼r)
<br>
=F
<br>
It doesnt depend upon value of r , so , if value of r is F then it also to be F so the condition exactly 2 are T is false . <strong>Hence option D is FALSE.</strong></p>Mathematical Logichttp://gateoverflow.in/1933/gate2014-1-53?show=133654#a133654Sun, 18 Jun 2017 05:36:36 +0000Answered: Testbook test series
http://gateoverflow.in/133514/testbook-test-series?show=133525#a133525
<p><a rel="nofollow" href="https://www.math.purdue.edu/~goldberg/Math453/burnside-probls.pdf">https://www.math.purdue.edu/~goldberg/Math453/burnside-probls.pdf</a></p>Mathematical Logichttp://gateoverflow.in/133514/testbook-test-series?show=133525#a133525Sat, 17 Jun 2017 07:01:20 +0000Answered: Testbook testseries
http://gateoverflow.in/133513/testbook-testseries?show=133524#a133524
Distinct Rings - each can be worn in either of 3 fingers<br />
<br />
, therefore, each ring = 3 ways<br />
<br />
total ways = 3 * 3 * 3 * 3 waysMathematical Logichttp://gateoverflow.in/133513/testbook-testseries?show=133524#a133524Sat, 17 Jun 2017 06:48:28 +0000[Self doubt] Permutation
http://gateoverflow.in/133489/self-doubt-permutation
Default permutation is with repetition or not?I came across some questions and it was nothing mentioned like repetition and all.But if questions says how many passwords then it was assumed that they are distinct and repetition not allowed and some questions asks like how many licenses number possible where 1 letter followed by 4 digits etc. and in these question it was with repitiion.So how to differentiate between these two?Mathematical Logichttp://gateoverflow.in/133489/self-doubt-permutationSat, 17 Jun 2017 00:10:37 +0000Answered: Rosen #MATHEMATICAL LOGIC
http://gateoverflow.in/133441/rosen-%23mathematical-logic?show=133475#a133475
there exist x,there exist y (x!=y ^love(lynn,x)^love(lynn,y)^¥z(love(lynn,z)->(z=x^z=y))).Mathematical Logichttp://gateoverflow.in/133441/rosen-%23mathematical-logic?show=133475#a133475Fri, 16 Jun 2017 18:58:36 +0000Answered: [Discrete Maths] predicate Logic
http://gateoverflow.in/133129/discrete-maths-predicate-logic?show=133324#a133324
<p>First convert this to have better understanding.
<br>
<strong>"No female like a male."
<br>
$\forall x$ $\forall y$ ( (Female(x) ^ Male(y)) $\rightarrow$ ~like(x, y) )</strong>
<br>
<br>
<strong>"No female like a male who doesn't like all vegetarian".
<br>
$\forall x$ $\forall y$ $\forall z$ ( (Female(x) ^ Male(y) ^ Veg(z) ^ ~like(y, z) ) $\rightarrow$ ~like(x, y) ).</strong>
<br>
<br>
$\forall x$ $\forall y$ $\forall z$ ( ~(Female(x) ^ Male(y) ^ Veg(z) ^ ~like(y, z) ) $\vee$ ~like(x, y) ).
<br>
As (P $\rightarrow$ Q) == (~P v Q)
<br>
Also ~(~P) = P. So we can put double negation, it can't affect.
<br>
<br>
$\forall x$ $\forall y$ $\forall z$ ~(~ ( ~(Female(x) ^ Male(y) ^ Veg(z) ^ ~like(y, z) ) $\vee$ ~like(x, y) ) ).
<br>
From here take one negation out of quantifiers and one negation inside statement. we can get this.
<br>
<strong> ~ ($\exists x$ $\exists y$ $\exists z$ ( (Female(x) ^ Male(y) ^ Veg(z) ^ ~like(y, z) ) ^ like(x, y)).
<br>
<br>
So ans: C</strong></p>Mathematical Logichttp://gateoverflow.in/133129/discrete-maths-predicate-logic?show=133324#a133324Fri, 16 Jun 2017 06:26:28 +0000Answered: # FREE VARIABLE ( NULL QUANTIFICATION ) Related problem
http://gateoverflow.in/132788/%23-free-variable-null-quantification-related-problem?show=133197#a133197
<a href="http://gateoverflow.in/132789/free-variable-and-validity-related-problem" rel="nofollow" target="_blank">http://gateoverflow.in/132789/free-variable-and-validity-related-problem</a>Mathematical Logichttp://gateoverflow.in/132788/%23-free-variable-null-quantification-related-problem?show=133197#a133197Thu, 15 Jun 2017 10:17:00 +0000Answered: FREE VARIABLE and VALIDITY related problem
http://gateoverflow.in/132789/free-variable-and-validity-related-problem?show=133193#a133193
<img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=9152570192656545297">
Caption
<p> </p>Mathematical Logichttp://gateoverflow.in/132789/free-variable-and-validity-related-problem?show=133193#a133193Thu, 15 Jun 2017 10:09:16 +0000Answered: [Discrete Maths] predicate logic
http://gateoverflow.in/132949/discrete-maths-predicate-logic?show=133190#a133190
Case 1: If for all the assignment to proposition variable if you get TRUE always then it is SATISFIABLE as well as TAUTOLOGY.<br />
<br />
Case 2: If for any assignment to proposition variable if you get TRUE then it is SATISFIABLE but not TAUTOLOGY.<br />
<br />
So statement "EVERY SATISFIABLE IS NOT TAUTOLOGY" TrueMathematical Logichttp://gateoverflow.in/132949/discrete-maths-predicate-logic?show=133190#a133190Thu, 15 Jun 2017 09:51:50 +0000Answered: GATE2014-3-53
http://gateoverflow.in/2087/gate2014-3-53?show=133183#a133183
Try this way<br />
<br />
NOT (all rainy days are cold)<br />
<br />
~(¥ d Rainy(d)->Cold(d))<br />
<br />
~(¥d ~Rainy(d) DISJUNCTION cold(d))<br />
<br />
∃d( Rainy (d) CONJUNCTION ~Cold(d))<br />
<br />
OPTION DMathematical Logichttp://gateoverflow.in/2087/gate2014-3-53?show=133183#a133183Thu, 15 Jun 2017 09:21:21 +0000Answered: GATE2004-IT-31
http://gateoverflow.in/3674/gate2004-it-31?show=133161#a133161
let p->q is conditional proposition here p and q are compound propositions itself<br />
Arguments to be valid if all combinations have to be tautology ( like T->T, F->T, F->F ) and its invalid if it have fallacy ( T->F )<br />
if we somehow get this fallacy (T->F) then an argument is invalid <br />
for options P and S u don't get any such combinations for T->F so P and S are VALID.<br />
for option Q : if we put p=F,q=T,r=T then we get T->F so it is INVALID <br />
for option R : if we put p=F,q=F,r=F then we get T->F so it is INVALID<br />
<br />
so Ans is C.Mathematical Logichttp://gateoverflow.in/3674/gate2004-it-31?show=133161#a133161Thu, 15 Jun 2017 06:57:28 +0000[Discrete Maths] First Order Logic
http://gateoverflow.in/133130/discrete-maths-first-order-logic
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=9721698226079299898"></p>Mathematical Logichttp://gateoverflow.in/133130/discrete-maths-first-order-logicThu, 15 Jun 2017 04:38:15 +0000Answered: GATE1990-3-x
http://gateoverflow.in/84861/gate1990-3-x?show=133093#a133093
To prove any wff valid or tautology try to use this analogy.<br />
<br />
Since implication A->B is FALSE only when A=T and B=F.So to prove any implication is valid or not try to get TRUE->FALSE if we succeed then it is not valid,if we not then wff is valid.<br />
<br />
So for option A<br />
<br />
substitute P=T and R=F<br />
<br />
RHS P->R become FALSE<br />
<br />
LHS (P->Q)^(P->R)<br />
<br />
To get true here we need T^T so substitute Q=T which makes P->Q TRUE and P->R FALSE so T^F=F which makes LHS=FALSE.<br />
<br />
Hence we are unable to get T->F which proves wff given in OPTION A is valid.<br />
<br />
NOTE: we can use similar kind of logic to prove contradiction.Mathematical Logichttp://gateoverflow.in/84861/gate1990-3-x?show=133093#a133093Wed, 14 Jun 2017 19:00:53 +0000Answered: Discrete Maths Graph theory
http://gateoverflow.in/132838/discrete-maths-graph-theory?show=133088#a133088
<p><em><strong>Euler Path:</strong></em></p>
<p><em>1.If the graph is connected and has exactly 2 odd degree vertices then there exists at least one Euler Path.Mainly it starts with one end and ends on the other End.</em></p>
<p>
<br>
<em><strong>Euler Circuit:</strong></em></p>
<p><em>1.If the graph is connected and every vertex has even degree then the Graph has atleast one Euler Circuit.</em></p>
<p><em>Euler Circuit is a part of Euler Path but Converse is nt True.</em></p>
<table border="2" cellpadding="0" cellspacing="0" style="width:80%">
<tbody>
<tr>
<td>
<p><em><strong># of ODD Vertices</strong></em></p>
</td>
<td>
<p><em><strong>Implication (for a connected graph)</strong></em></p>
</td>
</tr>
<tr>
<td>
<p><em>0</em></p>
</td>
<td>
<p><em>There is at least
<br>
one Euler Circuit.</em></p>
</td>
</tr>
<tr>
<td>
<p><em>1</em></p>
</td>
<td>
<p><em>THIS IS IMPOSSIBLE!</em>
<br>
</p>
</td>
</tr>
<tr>
<td>
<p><em>2</em></p>
</td>
<td>
<p><em>There is no Euler Circuit but at least 1 Euler Path.</em></p>
</td>
</tr>
<tr>
<td>
<p><em>more than 2</em></p>
</td>
<td>
<p><em>There are no Euler Circuits
<br>
or Euler Paths.</em></p>
</td>
</tr>
</tbody>
</table>Mathematical Logichttp://gateoverflow.in/132838/discrete-maths-graph-theory?show=133088#a133088Wed, 14 Jun 2017 18:20:42 +0000Answered: Counting
http://gateoverflow.in/132945/counting?show=132989#a132989
Let employees in office i = Oi<br />
<br />
Therefore : O1 + O2 +O3+ O4 = 6 where Oi>=1<br />
<br />
Putting at least 1 employee in each office we are left with (Oi complement = oi)<br />
<br />
o1 + o2 +o3+ o4 = 6 - 4 where oi>=0<br />
<br />
o1 + o2 +o3+ o4 = 2<br />
<br />
thus using combinations with repetition we have ( n +r -1 , r)<br />
<br />
n=4, r=2<br />
<br />
( 4 + 2 - 1 , 2 ) = (5,2) = 10 waysMathematical Logichttp://gateoverflow.in/132945/counting?show=132989#a132989Wed, 14 Jun 2017 08:14:54 +0000Answered: first order logic notations
http://gateoverflow.in/132898/first-order-logic-notations?show=132958#a132958
<p><strong>Bidirection</strong> ≡ <strong>Ex-Nor</strong> ≡ <strong>iff</strong> ≡ <strong>Equivalence operator</strong> ≡ <strong><=></strong></p>
<p>p<=><em>q </em>means (p=>q )AND(q>p) which means (p'+q).(q'+p) which means p'q'+pq</p>
<p>Now the prepositional logic p'q'+pq would be true when either both p,q are false or both are true.so this make p,q logically equivalent iff p'q'+pq is true always.</p>
<p>like for example let p=a+b' and q= (a'b)' so as you can guess p is logically equivalent to q here</p>
<p>but you can prove it by using p<=>q if the outcome for all possible inputs is always true then p and q would be logically equivalent.</p>
<p> </p>Mathematical Logichttp://gateoverflow.in/132898/first-order-logic-notations?show=132958#a132958Wed, 14 Jun 2017 03:29:50 +0000[Discrete maths ] Group theory
http://gateoverflow.in/132942/discrete-maths-group-theory
<p>Let G be abelian, H and K subgroups of G with orders n, m. Then G has subgroup of order lcm(n,m)</p>
<p>I have gone through prrof here:- <a rel="nofollow" href="https://math.stackexchange.com/questions/465742/let-g-be-abelian-h-and-k-subgroups-of-orders-n-m-then-g-has-subgrou">https://math.stackexchange.com/questions/465742/let-g-be-abelian-h-and-k-subgroups-of-orders-n-m-then-g-has-subgrou</a></p>
<p>But this uses assertion :- if G is abelian and n divides |G| then G has a subgroup of order n.How is this assertion true. there are some existing examples where n divides G but there is no subgroup with that order.Please clear with explanation.</p>
<p> </p>Mathematical Logichttp://gateoverflow.in/132942/discrete-maths-group-theoryTue, 13 Jun 2017 22:18:33 +0000[Discrete Maths] Hamilton Circuit,Rosen p-584,q47
http://gateoverflow.in/132840/discrete-maths-hamilton-circuit-rosen-p-584-q47
<p>How is the following graph is Hamilton?The answer says it has Hamilton circuit</p>
<p>.<img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=8250483095399749125"> </p>Mathematical Logichttp://gateoverflow.in/132840/discrete-maths-hamilton-circuit-rosen-p-584-q47Tue, 13 Jun 2017 02:12:09 +0000Discrete maths equivalence relation
http://gateoverflow.in/132834/discrete-maths-equivalence-relation
Consider the equivalence relation<br />
namely, R = (x, y) I x - y is an integer}.<br />
<br />
What is the equivalence class of R={ (x,y) | x-y in integer(Z)} equivalence<br />
relation?<br />
<br />
Given answer is :- (n + 1/2 I n E Z)<br />
<br />
My answer is :- (n*1/2 | n E Z)<br />
<br />
<br />
<br />
Is my answer correct also?Mathematical Logichttp://gateoverflow.in/132834/discrete-maths-equivalence-relationMon, 12 Jun 2017 20:41:16 +0000[Discrete Maths] Inclusion/Exclusion Probability
http://gateoverflow.in/132699/discrete-maths-inclusion-exclusion-probability
find the probability that when four numbers from 1 to 100,inclusive,are picked at random with no repetitions allowed,either all are odd,all divisible by 3,or all divisible by 5Mathematical Logichttp://gateoverflow.in/132699/discrete-maths-inclusion-exclusion-probabilitySun, 11 Jun 2017 17:56:46 +0000Answered: #rosen_chapter1
http://gateoverflow.in/132684/%23rosen_chapter1?show=132692#a132692
<p><strong>Answer of question a):</strong>
<br>
n-th statement: "Exactly n of the statements are false"
<br>
So, 1st statement: Exactly 1 of the statements are false
<br>
So, 2nd statement: Exactly 2 of the statements are false</p>
<p>.</p>
<p>.</p>
<p>each statement is contradictory to each other.if I say 1st statement is true i.e exactly 1 of the statement is false,so it means out of 100 statements 1 statement is false,correct?so,all 99 are correct statements. If rest 99 are correct, then let say, statement no. 3 which is "exactly 3 of the statements are false" should be correct, but if that statement no. 3 is correct then 3 statements are false,but according to 1st statement that is wrong,similarly u can see that, all 100 statements are contradict within themselves,so maximum one can be true, and thats possible when only statement no. 99 is true, that is "exactly 99 of the statements are false" and yeah thats true and only 1 statement is correct which is statement no. 99. So,all statements no. (1-98) and 100 are wrong,and 99 is only correct.
<br>
<strong>Answer of question b) Part1:</strong></p>
<p> if the <em>n</em>th statement is “At least <em>n </em>of the statements in this list are false.”</p>
<p>statement 1: At least 1<em> </em>of the statements in this list are false.</p>
<p>Statement 2: At least 2<em> </em>of the statements in this list are false.</p>
<p>.</p>
<p>.</p>
<p>So if i say statement 1 is false, then it means that no statement is false,but I am saying that statement 1 is false,its again conflicting,so statement 1 must not be false,it should be true,</p>
<p>Similarly for statement 2,if i say it is false,then it means, no of false statements is<2 i.e 1, but we have already declared that statement 1 must be true,so again statement 2 need to be true.
<br>
Same thing happens for statement no.3,4,5....,50, because,</p>
<p>For statement 50, it says:"At least 50<em> </em>of the statements in this list are false.", I have already declared that 1-49 are true,so if statement no. 50 is false,then no. of false statements<50, but already 49 statements are true and 0 statements are false,so statement no. 50 must also be true.
<br>
For statement no 51, if i say it is true then,atleast 51 statements have to be false, but already 50 statements are true and 50 are remaining to be checked,so statement no. 51 must not be true,so it should be false,so if it is false it means no of false statements<51,and its matching with our logic,because already 50 are true,so rest are all false,so 1-50 are true and 51-100 are false.</p>
<p><strong>Answer of question b) Part 2:</strong></p>
<p>In a same manner, what we have seen just above 1-49 is true,</p>
<p>What about statement no. 50?</p>
<p>If it is true, there must be at least 50 false statements. As 1−50 are true in this case, this can never be the case as we only have 99 statements</p>
<p>If it is false, it is not the case that at least 50 of the statements are false. As 1−49 are true, this can only be the case if at least one of the statements in the interval 52−99 is true. This can never be the case for the above reason. Thus statement 51 is neither true or false.
<br>
So its a paradox.</p>Mathematical Logichttp://gateoverflow.in/132684/%23rosen_chapter1?show=132692#a132692Sun, 11 Jun 2017 16:54:17 +0000Answered: #rosen_chapter1
http://gateoverflow.in/132683/%23rosen_chapter1?show=132688#a132688
<p>follow this link:
<br>
<a rel="nofollow" href="https://www.facebook.com/groups/gateoverflow/permalink/643620245843189/">https://www.facebook.com/groups/gateoverflow/permalink/643620245843189/</a></p>Mathematical Logichttp://gateoverflow.in/132683/%23rosen_chapter1?show=132688#a132688Sun, 11 Jun 2017 16:10:42 +0000Answered: #rosen_chapter1
http://gateoverflow.in/132682/%23rosen_chapter1?show=132685#a132685
Yeah correct.Mathematical Logichttp://gateoverflow.in/132682/%23rosen_chapter1?show=132685#a132685Sun, 11 Jun 2017 16:05:56 +0000[Discrete maths] Generating function
http://gateoverflow.in/132616/discrete-maths-generating-function
9. Find the coefficient of x lO in the power series of each of<br />
these functions.<br />
a. (x^2 +x^4 +x^6 +x^8 + .. . )(x^3 +x^6 +x^9 + .. ·)(x^4 +<br />
x^8 +X^12 + ... )<br />
b. (I +x^2+x^4+x^6+ x^8+ ... )(1 +x^4+X^8+X^I2 + ... )<br />
(I +x^6 +x^12 +x^18 + ... )Mathematical Logichttp://gateoverflow.in/132616/discrete-maths-generating-functionSun, 11 Jun 2017 04:26:43 +0000Answered: [Discrete maths] graph theory Perfect matching
http://gateoverflow.in/132353/discrete-maths-graph-theory-perfect-matching?show=132609#a132609
<p><strong>Assumptions:</strong></p>
<ul>
<li>by 'graph' I mean a simple graph.</li>
<li>$n$ = no. of vertices in a graph.</li>
<li>$e$ = no. of edges in a graph.</li>
</ul>
<p><strong>Definitions:</strong></p>
<ul>
<li>Matching - it is a set of non-adjacent edges of a graph.</li>
<li>Perfect Matching - it is a matching which matches every vertex with some other vertex in the graph. Perfect matching can't exist for graphs containing odd no. of vertices, because one unmatched vertex will always be left.</li>
<li>Matching Number - it is the size of the largest matching of a graph.</li>
<li>Covering (or Edge Cover or Line Cover) - it is a set of edges (adjacent or non-adjacent) such that every vertex of the graph is covered by at least one edge in that set. Covering can't exist for graph containing one or more isolated vertices.</li>
<li>Covering Number - it is the size of the smallest covering of a graph.</li>
</ul>
<p><strong>Answer:</strong></p>
<p>You think right. It is not possible for a graph to have the same Matching and Covering numbers without having a Perfect Matching.</p>
<p>The points (1) and (2) below, show that the Matching Number and Covering Number of a graph can be same only when both are equal to $\frac{n}{2}$. And, the points (3) and (4) show that if the Matching Number and Covering Number of a graph are both equal to $\frac{n}{2}$, then there will always exist AT LEAST one set of edges which is both a Perfect Matching as well as a Covering.</p>
<ol>
<li>When an edge joins two vertices, the two vertices of the edge are matched to each other.
<ul>
<li>What can be the minimum no. of edges in a Matching ?
<ul>
<li>A set containing a single edge is always a matching as there is no other edge in the set to which it can be adjacent to.</li>
</ul>
</li>
<li>What can be the maximum no. of edges in a Matching ?
<ul>
<li>A matching set can have a maximum of $\left \lfloor \frac{n}{2} \right \rfloor$<strong><sup>a</sup></strong> non-adjacent edges because if even one more edge is further added then that edge will become adjacent to some edge already present in the set, and the set will no more be a matching.</li>
</ul>
</li>
<li>Therefore, $1$ $\leq$ $Matching$ $Number$ $\leq \left \lfloor \frac{n}{2} \right \rfloor$</li>
</ul>
</li>
<li>Similarly, when an edge joins two vertices, both the vertices are said to be covered.
<ul>
<li>What can be the maximum no. of edges in a Covering ?
<ul>
<li>Since there is no restriction on the adjacency of edges in a Covering and the edge set (of a graph with no isolated vertices) always covers all the vertices, the edge set of a graph is always a Covering. Hence, the size of a Covering can't exceed $e$.</li>
</ul>
</li>
<li>What can be the minimum no. of edges in a Covering ?
<ul>
<li>Since a single edge covers only two vertices and there are a total of $n$ vertices to be covered, therefore a minimum of $\left \lceil \frac{n}{2} \right \rceil$<strong><sup>b</sup></strong> edges is necessary to form a Covering.</li>
</ul>
</li>
<li>Therefore, $\left \lceil \frac{n}{2} \right \rceil \leq$ $Covering$ $Number$ $\leq e$.</li>
</ul>
</li>
<li>In a graph where $n$ is even, if there exists a Matching of size $\frac{n}{2}$, then the Matching Number of the graph is $\frac{n}{2}$ because it can't exceed $\frac{n}{2}$. And, the Matching itself becomes a Perfect Matching because the $\frac{n}{2}$ non-adjacent edges in the matching, together, match and pair every vertex of the graph with some other vertex of the graph. Also, since a Perfect Matching matches every vertex with some vertex, all the vertices of the graph are covered. Therefore, every Perfect Matching is a Covering.</li>
<li>If a Covering contains $\frac{n}{2}$ edges, and $n$ is even, then the Covering number of the graph is $\frac{n}{2}$ as it can't be smaller any more. Also, the $\frac{n}{2}$ edges of such a Covering will always be non-adjacent to each other, because otherwise (as shown in the below figure) it wouldn't then be possible to cover all the $n$ vertices and at least one vertex would be left uncovered. Therefore, given that $n$ is even, every Covering of $\frac{n}{2}$ edges is a Perfect Matching.</li>
</ol>
<p style="text-align:center"><img alt="" height="213" src="http://gateoverflow.in/?qa=blob&qa_blobid=1609986996340351664" width="400"></p>
<hr>
<p><strong><sup>a</sup></strong> The maximum no. of edges possible in a Matching is "floor" of $\frac{n}{2}$ because if there are even no. of vertices then all of them can be matched using $\frac{n}{2}$ non-adjacent edges, but if there are odd no. of vertices then only $n-1$ vertices can be matched and paired using $\frac{n-1}{2}$ non-adjacent edges. So, we generalize and combine both the cases by saying that the max. possible no. of edges in a Matching is $\left \lfloor \frac{n}{2} \right \rfloor$.</p>
<p><strong><sup>b</sup></strong> The minimum no. of edges possible in a Covering is "ceiling" of $\frac{n}{2}$ because if there are even no. of vertices then $\frac{n}{2}$ edges can cover all of them, but if there are odd no. of vertices then only $n-1$ vertices can be covered using $\frac{n-1}{2}$ edges. One more edge will be needed to cover the remaining uncovered vertex, thus requiring a total of $\frac{n-1}{2}+1=\frac{n+1}{2}$ edges for the Covering. So, to generalize, we say that min. possible no. of edges in a Covering is $\left \lceil \frac{n}{2} \right \rceil$.</p>Mathematical Logichttp://gateoverflow.in/132353/discrete-maths-graph-theory-perfect-matching?show=132609#a132609Sun, 11 Jun 2017 00:40:20 +0000Answered: Gatebook - predicate Logic - negating
http://gateoverflow.in/132090/gatebook-predicate-logic-negating?show=132608#a132608
Let L(x,y) represent x loves y<br />
<br />
1. ∀x∀yL(x, y)) :Given predicate<br />
<br />
Negation ¬∀x∀yL(x, y) = ∃x∃y¬L(x, y) : There is someone who doesn't love someone.Somebody hates somebody<br />
<br />
2. ∀x∃y¬L(x, y) :Given predicate says everyone hates someone<br />
<br />
Negation ¬∀x∃y¬L(x, y) =∃x∀yL(x, y) :- there is someone who loves everyone.<br />
<br />
3.∃x∃yL(x, y) <br />
<br />
Negation ∀x∀y¬L(x, y) : Everyone hates everyone. or nobody loves somebody<br />
<br />
4.∀x∃yL(x, y) <br />
<br />
Negation:- .∃x∀y¬L(x, y) : There is someone who hates everyone.Or somebody loves nobodyMathematical Logichttp://gateoverflow.in/132090/gatebook-predicate-logic-negating?show=132608#a132608Sat, 10 Jun 2017 23:04:02 +0000Answered: Mathematical Logic
http://gateoverflow.in/132576/mathematical-logic?show=132602#a132602
1.Are => and -> used in the same way?<br />
<br />
yes.its just a way of representation.Arrow is in one direction is one way implication.Dont get confuse between single and double line.<br />
<br />
2.=> is logical implication and <=> is equivalence right?<br />
<br />
Yes.=> is logical implicatin.it implies in one direction ,while <==> implies both ways.<br />
<br />
3.Then why does in some questions, => and <=> is read as 'if then'? '->' symbol is for if then right?<br />
<br />
a:)If p then q is same as p->q ,it does not mean then q->p.its goes one way.One is symbolic representation,other is English.<br />
<br />
b:) If p then q is not p<==>q. .<br />
<br />
Refer rosen book.In first chapter they have mentioned around 15 ways of different representation of these symbols.Mathematical Logichttp://gateoverflow.in/132576/mathematical-logic?show=132602#a132602Sat, 10 Jun 2017 21:10:51 +0000[Discrete maths] Permutations and combinations
http://gateoverflow.in/132599/discrete-maths-permutations-and-combinations
<pre>
How many ways can n books be placed on k distinguishable shelves
a. if no two books are the same ,and the positions of the books on the shelves does not matter?
b. if no two books are the same,and the positions of the books on the shelves matter?
</pre>Mathematical Logichttp://gateoverflow.in/132599/discrete-maths-permutations-and-combinationsSat, 10 Jun 2017 20:38:11 +0000