By QB365 on 02 Sep, 2022
QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Maths Subject - Discrete Mathematics, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
12th Standard
Maths
Verify the
(i) closure property,
(ii) commutative property,
(iii) associative property
(iv) existence of identity and
(v) existence of inverse for the arithmetic operation + on Z.
Verify the
(i) closure property,
(ii) commutative property,
(iii) associative property
(iv) existence of identity and
(v) existence of inverse for the arithmetic operation - on Z.
Verify the
(i) closure property,
(ii) commutative property,
(iii) associative property
(iv) existence of identity and
(v) existence of inverse for the arithmetic operation + on Ze = the set of all even integers
Verify the
(i) closure property,
(ii) commutative property,
(iii) associative property
(iv) existence of identity and
(v) existence of inverse for the arithmetic operation + on Zo = the set of all odd integers
Verify
(i) closure property
(ii) commutative property, and
(iii) associative property of the following operation on the given set. (a*b) = ab;∀a, b∈N (exponentiation property)
Verify
(i) closure property
(ii) commutative property
(iii) associative property
(iv) existence of identity, and
(v) existence of inverse for following operation on the given set m*n = m + n - mn; m, n ∈Z
Verify
(i) closure property
(ii) commutative property
(iii) associative property
(iv) existence of identity and
(v) existence of inverse for the operation +5 on Z5 using table corresponding to addition modulo 5.
Verify
(i) closure property
(ii) commutative property
(iii) associative property
(iv) existence of identity and
(v) existence of inverse for the operation ×11 on a subset A = {1, 3, 4, 5, 9} of the set of remainders {0,1, 2, 3, 4, 5, 6, 7, 8, 9,10}
Let p: Jupiter is a planet and q: India is an island be any two simple statements. Give verbal sentence describing each of the following statements.
(i) ¬p
(ii) p ∧ ¬q
(iii) ¬p ∨ q
(iv) p➝ ¬q
(v) p↔q
Define an operation \(*\)on Q as follows: a * b =\(\left( \frac { a+b }{ 2 } \right) \); a,b ∈Q. Examine the closure, commutative, and associative properties satisfied by \(*\)on Q.
Define an operation∗ on Q as follows: a*b = \(\left( \frac { a+b }{ 2 } \right) \); a,b ∈Q. Examine the existence of identity and the existence of inverse for the operation * on Q.
Verify whether the following compound propositions are tautologies or contradictions or contingency
(p ∧ q) ∧ ¬ (p ∨ q)
Verify whether the following compound propositions are tautologies or contradictions or contingency
(( p V q)∧ ¬ p) ➝ q
Verify whether the following compound propositions are tautologies or contradictions or contingency
( p ⟶ q) ↔️ (~ p ⟶ q)
Verify whether the following compound propositions are tautologies or contradictions or contingency
((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r)
Let M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and let * be the matrix multiplication. Determine whether M is closed under ∗. If so, examine the commutative and associative properties satisfied by ∗ on M.
Let M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and let ∗ be the matrix multiplication. Determine whether M is closed under ∗ . If so, examine the existence of identity, existence of inverse properties for the operation ∗ on M.
Let A be Q\{1}. Define ∗ on A by x*y = x + y − xy. Is ∗ binary on A? If so, examine the commutative and associative properties satisfied by ∗ on A.
Check whether the statement p➝(q➝p) is a tautology or a contradiction without using the truth table.
Using truth table check whether the statements ¬(p V q) V (¬p ∧ q) and ¬p are logically equivalent.
Let A be Q\{1}. Define ∗ on A by x*y = x + y − xy . Is ∗ binary on A? If so, examine the existence of identity, existence of inverse properties for the operation ∗ on A.
Prove p⟶(q⟶r) ☰ (p ∧ q)⟶r without using truth table.
Prove that p➝(¬q V r) ≡ ¬pV(¬qVr) using truth table.
Show that \(\neg(p \wedge q) \equiv \neg p \vee \neg q\)
Show that \(\neg(p \rightarrow q) \equiv p \wedge \neg q\)
Answers
(i) m + n∈Z, ∀m, n∈Z. Hence + is a binary operation on Z.
(ii) Also m + n = n + m,∀m, n∈Z. So the commutative property is satisfied
(iii) ∀m, n, p∈Z, m+ (n + p) = (m+ n) + p. Hence the associative property is satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ヨ 0∈Z⋺(m+ 0) = (0 + m) = m. Hence the existence of identity is assured.
(v) m + m' = m'+ m = 0 ⇒ m' = −m. Thus ∀∈Z,ョ−m∈Z ⋺ m+ (−m) = (−m) + m = 0. Hence, the existence of inverse property is also assured. Thus we see that the usual addition + on Z satisfies all the above five properties.
i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6 …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).
Consider the set of all even integers Ze = {2k | k ∈ Z} = {...,−6, −4, −2, 0, 2, 4, 6,...}.
Let us verify the properties satisfied by + on Ze.
(i) The sum of any two even integers is also an even integer.
Because x, y∈Ze, ⇒ x = 2m and y = 2n , m,n∈Z.
So (x + y) = 2m + 2n = 2(m+n)∈Ze. Hence + is a binary operation on Ze.
(ii) ∀ x, y∈Ze, (x + y) = 2(m + n) = 2(m + n) = 2(n + m) = (2n + 2m) = (y + x).
So + has commutative property
(iii) Similarly it can be seen that ∀x, y, z∈Ze, (x + y) + z = x + ( y + z).
Hence the associative property is true.
(iv) Now take x = 2k , then 2k + e = e + 2k = 2k ⇒ e = 0.
Thus ∀ x ∈ Ze, ヨ0∈Ze, ⋺x+0 = 0+x = x.
So, 0 is the identity element.
(v) Taking x = 2k and x′ as its inverse, we have 2k+x' = 0 = x'+2k ⇒ x' = −2k. i.e., x' = −x.
Thus ∀x ∈ Ze, ヨ-x∈Ze ⋺x + (−x) = (−x) + x = 0
Hence -x is the inverse of x ∈Ze.
Consider the set Zo of all odd integers Zo = {2k+1 : k∈Z} = {.., -5,-3,-1,1,3,5,...}. + is not a binary operation on Zo because when x = 2m+1, y = 2n +1, x + y = 2(m+ n) + 2 is even for all m and n. For instance, consider the two odd numbers 3, 7 ∈Zo. Their sum 3 + 7 = 10 is an even number. In general, if x, y∈Z0, then (x+y)∉Zo. Other properties need not be checked as it is not a binary operation.
(i) It is true that a*b = ab ∈N; ∀a,b∈N. So * is a binary operation on N.
(ii) a*b = ab and b*a = ba. Put, a = 2 and b = 3. Then a*b = 23 = 8 but b*a = 32 = 9 So a*b need not be equal to b*a. Hence * does not have commutative property.
(iii) Next consider a*(b*c) = a*(bc) = a(bc) Take a = 2, b = 3 and c = 4.
Then a*(b*c) = 2*(*4) = 234 = 281
But (a*b)*c = (ab)*c = (ab)c = a(bc) = a(bc) = 212
Hence a*(b*c) ≠ (a*b)*c. So * does not have associative property on N.
(i) The output m+ n - mn is clearly an integer and hence∗ is a binary operation on Z.
(ii) m*n = m+ n − mn = n + m − nm = n*m, ∀m,n∈Z. So ∗ has commutative property.
(iii) Consider (m*n)*p = (m+ n −m n)* p= (m+ n −mn) + p − (m+ n −m n) p
= m+ n + p −mn −m p − n p + m n p ... (1)
Similarly m*(n*p) = m*(n + p − n p) = m+ (n + p − n p) −m (n + p − n p)
= m+ n + p − n p −m n −mp + m n p ... (2)
From (1) and (2), we see that m*(n*p) = (m*n)*p. Hence * has associative property.
(iv) An integer e is to be found such that
m*e = e*m = m, ∀m∈Z ⇒ m + e - m e = m
⇒e(1-m) = 0 ⇒ e = 0 or m = 1. But m is an arbitrary integer and hence need not be equal to 1. So the only possibility is e = 0. Also m*0 = 0*m = m, ∀m∈Z. Hence 0 is the identity element and hence the existence of identity is assured.
(v) An element m'∈ Z is to be found such that m*m' = m' * m = e = 0, ∀m∈Z.
m*m' = 0 ⇒ m+m'-m m'= 0⇒ m m' = 0 ⇒ \(\frac{m}{m-1}\). when m = 1, m' is not defined.
When m = 2, m' is an integer. But except m = 2, m′ need not be an integer for all values of m. Hence inverse does not exist in Z.
It is known that Z5 = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.
+5 | 0 | 1 | 2 | 3 | 4 |
0 | 0 | 1 | 2 | 3 | 4 |
1 | 1 | 2 | 3 | 4 | 0 |
2 | 2 | 3 | 4 | 0 | 1 |
3 | 3 | 4 | 0 | 1 | 2 |
4 | 4 | 0 | 1 | 2 | 3 |
(i) Since each box in the table is filled by exactly one element of Z5, the output a +5 b is unique and hence +5 is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +5 has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+53)+5 4 = 0+5 4 = 4(mod 5)
and 2+5(3+54) = 2 +5 2 = 4(mod5)
Hence (2+53)+54 = 2+5(3+54)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +5 is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z5, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+5 5+ = 0 (mod5). In this way, the inverse of each and every element of Z5 can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.
The table for the operation x11 is as follows.
x11 | 1 | 3 | 4 | 5 | 9 |
1 | 1 | 3 | 4 | 5 | 9 |
3 | 3 | 9 | 1 | 4 | 5 |
4 | 4 | 1 | 5 | 9 | 3 |
5 | 5 | 4 | 9 | 3 | 1 |
9 | 9 | 5 | 3 | 1 | 4 |
Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×11 on A is given below.
(i) Since each box has an unique element of A, ×11 is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×11 has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×11. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.
Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.
Given \(a*b=\frac { a+b }{ 2 } \forall \in Q\)
i) Closure property:
Let a, b ∈ Q
∴ a*b = \(\frac{a+b}{2}\)∈Q
[∵ addition and division are closed on Q]
* is closed on Q.
(ii) Commutative property:
Let a, b ∈ Q
Then a+b \(=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } =b*a\)
∴ a*b = b*a ∀a,b∈Q
∴ * is commutative on Q.
(iii) Associative property :
Let a, b, c ∈ Q
a*(b*c) = (a*b)*c
Let a = 2, b = 3, c-5
∴ a*(b*c) = 2*(3*-5)
\(=*\left( \frac { 3-5 }{ 2 } \right) \)
\(=2*(-1)=\frac { 2+(-1) }{ 2 } \)
\(=\frac { 1 }{ 2 } \quad \quad ...(1)\)
Now (a*b)*c = (2*3)*(-5)
\(=\left( \frac { 2+3 }{ 2 } \right) *(-5)\)
\(=\frac { 5 }{ 2 } *-5=\frac { \frac { 5 }{ 2 } +(-5) }{ 2 } \)
\(=\frac { 5-10 }{ 4 } =\frac { -5 }{ 4 } ...(2)\)
From (1) & (2), a*(b*c) ≠ (a*b)*c
∴ * is not associative on Q.
Given \(a*b=\frac { a+b }{ 2 } \), where a.b ∈Q Let a,b ∈Q
An element e has to found out such that
a*e = e*a = a
Let a = 5, Then 5*e = 5
\(\Rightarrow \frac { 5+e }{ 2 } =\)5 ⇒ 5 + e = 10
Let a = \(\frac{2}{3}\). Then \(\frac{2}{3}\)*e = \(\frac{2}{3}\)
\(\Rightarrow \frac { \frac { 2 }{ 3 } +e }{ 2 } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2 }{ 3 } +e=\frac { 4 }{ 3 } \)
\(\Rightarrow e=\frac { 4 }{ 3 } -\frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
It is seen that for the binary operation * defined on Q, identity element e is not unique. Hence identity element not defined for the binary operation * on Q.
The identity does not exist. Hence inverse also does not exist for the operation * on Q.
p | q | p ∧ q | p ∨ q | ~p ∨ q | (p ∧ q) ∧ ~(p ∨ q) |
T | T | T | T | F | F |
T | F | F | T | F | F |
F | T | F | T | F | F |
F | F | F | F | T | F |
The statement (p ∧ q) ∧ ~(p ∨ q)
(( p V q)∧ ~p)) ➝ q
p | q | p V q | ~p | ( p V q) ∧ ~q | ( p V q) ∧ ~q |
T | T | T | F | F | T |
T | F | T | F | F | T |
F | T | T | T | T | T |
F | F | F | T | F | T |
The statement (( p V q)∧ ~p) ➝ q is a tautology.
p | q | p ⟶ q | ~p | ~p ⟶ q | ( p ⟶ q) ↔️ (~p ⟶ q) |
T | T | T | F | T | T |
T | F | F | F | T | F |
F | T | T | T | T | T |
F | F | T | T | F | F |
Since this is neither a tautology not a contradiction
( p ⟶ q) ↔️ (~p ⟶ q) is a contingency.
p | q | r | q ⟶ r | p⟶ q | q ⟶ r | (p⟶ q) ∧ (q ⟶ r) | ((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r) |
T | T | T | T | T | T | T | T |
T | T | F | F | T | F | F | T |
T | F | T | T | F | T | F | T |
T | F | F | F | F | T | F | T |
F | T | T | T | T | T | F | T |
F | T | F | T | T | F | F | T |
F | F | T | T | T | T | T | T |
F | F | F | T | T | T | T | T |
∴ ((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r)
The given statement is a tautology.
Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and * be the matrix multipilication.
Let A \(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \) and B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \in M\)
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(\\ =\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \in M\)
[∵ 2xy ∈R-{0}]
∴ M is closed under *.
Commutative property:
we know A*B =\(\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) ..(1)\)
Let x,y∈R-{0}
Now B + A \(=\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \\ \)
From (1) &(2), A*B = B*A
∴ *has commutative property on M
Associative property:
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
B =\(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \) and
C = \(\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
for x, y, z ∈R-{0}
\((A*B)*C=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) *\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xyz+2xyz & 2xyz+2xyz \\ 2xyz+2xyz & 2xyz+2xyz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(1)\)
Now\(A*(B*C)=A*\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) *\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(2)\\ \)
From (1)&(2), (a*B)*C = A*B*C)
Since matrix multiplication is associative, this axiom holds good for M.
Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and ∗ be the matrix multiplication.
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)and
B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)∈M
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & xy \\ 2xy & 2xy \end{matrix} \right) \in M\\ \)
[∵ 2xy∈R-{0}]
∴ M is closed under M
Identity:
Since identity of 2\(\times\)2 matrices is I =\(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \)∉M
∴ M has no identity under *.
Inverse:
Since it has no identity, it won't have inverse also.
given A = {Q\{1}}
A is defined on A by x*y = x+y-xy
Let x,y ≠ 1
∴ x*y = x + y - xy
Now to prove that x + y - xy ≠ 1
Let us assume that x + y - xy = 1
x+y-xy-1 = 0
(x-1)-y(x-1) = 0
(x-1)(1-y) = 0
x =1 or y = 1 which is a false [∵x, y ≠ 1]
∴ is a binary operation on A.
Commutative property:
Let x,y ∈A ⇒ x, y≠1
∴x*y = x+y-xy
and y*x = y+x-yx
⇒x+y = y*x∀x, y∈A
A has commutative property under *
Associative property:
Let x,y,z ∊A ⇒x,y,z≠1
Consider (x*y)*z = (x+y-xy)*z
= x +y~xy +z- (x +y-xy)z
= x +y-xy+ z-xz- yz + xyz
= x +y+z-xy-yz-zx +xyz ...(1)
= x +y+z - yz - x (y + z - yz)
= x +y +z-yz-zy-xz +xyz ..(2)
From (1) & (2), (x*y)*z = x*(y*z)
A has associative property under *.
Given statement is p ➝(q ➝ p)
=p ➝ (~q v p)
[∴ By example p ➝ q =p V q]
= (~p) V (~q V p) [Again by example 12.17]
= (~p V ~q) V p [Associative law)]
= (~p V p) V ~q [Commutative law)]
= II V ~q [~p V p is also tautology]
= II [II V ~q] is also a tautology.
~(p V q) V (~p ∧ q) and ~p
p | q | p V q | ~(p ∧ q) | ~p | ~p ∧ q | ~(p V q) V (~p ∧ q) |
T | T | T | F | F | F | F |
T | F | T | F | F | F | F |
F | T | T | F | T | T | T |
F | F | F | T | T | F | T |
The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.
Existence of identity:
We have to find an element a' ∈ A such that
a*a' = a'*a = e
⇒ a + e-ae = a
⇒ a + e-ae = a
⇒ e-ae = 0
⇒ e(1-a) = 0
⇒ e = \(\frac{0}{1-a}\) = 0∈A
∴ A has an identity under *.
Existence of inverse:
For every a ∈ A, there exist a' ∈ A such that a*a' = a' * a = e
⇒ a+a'-aa' = 0 [∵e=0]
⇒ a+a'(1-a) = 0
⇒ a'(1-a) = -a
⇒ a' = \(\frac{-a}{1-a}\)
To prove that \(\frac{-a}{1-a}\)≠ 1
Suppose \(\frac{-a}{1-a}\) = 1
⇒ -a = 1-a
⇒ -a + a = 1 ⇒ 0 ≠ 1
∴ Our assumption is wrong
⇒ \(\frac{-a}{1-a}\) ≠ 1
∴ A has inverse for every element x ∈ A
Prove that p⟶(q⟶r) = (p ∧ q)⟶r without using truth table. From example we know that p⟶ q = ~p V q
Consider LHS = p⟶(q⟶r)
= p ⟶ (~q V r) [Implication Law]
= ~p V (~q V r) [Implication Law]
= (~p V ~q) V r [associative property]
= ~(p ∧ q) V r [using Demorgan's law]
= (p ∧ q) ⟶ r [Implication Law]
Hence proved.
p | q | r | ~ q | ~q V r | p➝(¬qVr) | ~p | ~pV(~qVr) |
T | T | T | F | T | T | F | T |
T | T | F | F | F | F | F | F |
T | F | T | T | T | T | F | T |
T | F | F | T | T | T | F | T |
F | T | T | F | T | T | T | T |
F | T | F | F | F | T | T | T |
F | F | T | T | T | T | T | T |
F | F | F | T | T | T | T | T |
From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴ p➝(¬q V ~r) ≡ ~pV(~q V r)
Hence proved.
Truth table for \(\neg(p \wedge q)\)
P | Q | \(p \wedge q\) | \(\neg(p \wedge q)\) |
T | T | T | F |
T | F | F | T |
F | T | F | T |
F | F | F | T |
Truth table for \(\neg p \vee \neg q\)
P | Q | \(\neg p\) | \(\neg q\) | \(\neg p \vee \neg q\) |
T | T | F | F | F |
T | F | F | T | T |
F | T | T | F | T |
F | F | T | T | T |
To prove \(\neg(p \rightarrow q) \equiv p \wedge \neg q\)
Truth table for \(\neg(p \rightarrow q)\)
p | q | \(p \rightarrow q \) | \(\neg(p \rightarrow q)\) |
T | T | T | F |
T | F | F | T |
F | T | T | F |
F | F | T | F |
Truth table for \(p \wedge \neg q\)
p | q | \(\neg q\) | \(p \wedge \neg q\) |
T | T | F | F |
T | F | T | T |
F | T | F | F |
F | F | T | T |
The entries in the column \(\neg(p \rightarrow q) \text { and } p \wedge \neg q\) are identical and they are equivalent.