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Webpage for Mathematical Logic

Important Question Types:

  1. Checking validity of First Order Logic Statements
  2. Checking validity of Propositional Logic

Recent questions tagged mathematical-logic

0 votes
1 answer
1
TIFR2021-B: 1
Consider the following statements about propositional formulas. $\left ( p\wedge q \right )\rightarrow r$ and $\left ( p \rightarrow r \right )\wedge \left ( q\rightarrow r \right )$ are $\textit{not }$ ... on the values $p$ and $q$, $\text{(i)}$ can be either true or false, while $\text{(ii)}$ is always false.
Consider the following statements about propositional formulas. $\left ( p\wedge q \right )\rightarrow r$ and $\left ( p \rightarrow r \right )\wedge \left ( q\rightarrow r \right )$ are $\textit{not }$ ... Depending on the values $p$ and $q$, $\text{(i)}$ can be either true or false, while $\text{(ii)}$ is always false.
asked Mar 25 in Others soujanyareddy13 74 views
3 votes
3 answers
2
GATE CSE 2021 Set 2 | Question: 15
Choose the correct choice(s) regarding the following proportional logic assertion $S$: $S: (( P \wedge Q) \rightarrow R) \rightarrow (( P \wedge Q) \rightarrow (Q \rightarrow R))$ $S$ is neither a tautology nor a contradiction $S$ is a tautology $S$ is a contradiction The antecedent of $S$ is logically equivalent to the consequent of $S$
Choose the correct choice(s) regarding the following proportional logic assertion $S$: $S: (( P \wedge Q) \rightarrow R) \rightarrow (( P \wedge Q) \rightarrow (Q \rightarrow R))$ $S$ is neither a tautology nor a contradiction $S$ is a tautology $S$ is a contradiction The antecedent of $S$ is logically equivalent to the consequent of $S$
asked Feb 18 in Mathematical Logic Arjun 806 views
1 vote
4 answers
3
GATE CSE 2021 Set 1 | Question: 7
Let $p$ and $q$ be two propositions. Consider the following two formulae in propositional logic. $S_1: (\neg p\wedge(p\vee q))\rightarrow q$ $S_2: q\rightarrow(\neg p\wedge(p\vee q))$ Which one of the following choices is correct? Both $S_1$ and ... but $S_2$ is not a tautology $S_1$ is not a tautology but $S_2$ is a tautology Neither $S_1$ nor $S_2$ is a tautology
Let $p$ and $q$ be two propositions. Consider the following two formulae in propositional logic. $S_1: (\neg p\wedge(p\vee q))\rightarrow q$ $S_2: q\rightarrow(\neg p\wedge(p\vee q))$ Which one of the following choices is correct? Both $S_1$ and $S_2$ are tautologies ... tautology but $S_2$ is not a tautology $S_1$ is not a tautology but $S_2$ is a tautology Neither $S_1$ nor $S_2$ is a tautology
asked Feb 18 in Mathematical Logic Arjun 608 views
1 vote
2 answers
4
UGCNET-Oct2020-II: 3
Which of the following pairs of propositions are not logically equivalent? $((p \rightarrow r) \wedge (q \rightarrow r))$ and $((p \vee q) \rightarrow r)$ $p \leftrightarrow q$ and $(\neg p \leftrightarrow \neg q)$ $((p \wedge q) \vee (\neg p \wedge \neg q))$ and $p \leftrightarrow q$ $((p \wedge q) \rightarrow r)$ and $((p \rightarrow r) \wedge (q \rightarrow r))$
Which of the following pairs of propositions are not logically equivalent? $((p \rightarrow r) \wedge (q \rightarrow r))$ and $((p \vee q) \rightarrow r)$ $p \leftrightarrow q$ and $(\neg p \leftrightarrow \neg q)$ $((p \wedge q) \vee (\neg p \wedge \neg q))$ and $p \leftrightarrow q$ $((p \wedge q) \rightarrow r)$ and $((p \rightarrow r) \wedge (q \rightarrow r))$
asked Nov 20, 2020 in Discrete Mathematics jothee 671 views
0 votes
1 answer
5
UGCNET-Oct2020-II: 37
If $f(x)=x$ is my friend, and $p(x) =x$ is perfect, then correct logical translation of the statement “some of my friends are not perfect” is ______ $\forall _x (f(x) \wedge \neg p(x))$ $\exists _x (f(x) \wedge \neg p(x))$ $\neg (f(x) \wedge \neg p(x))$ $\exists _x (\neg f(x) \wedge \neg p(x))$
If $f(x)=x$ is my friend, and $p(x) =x$ is perfect, then correct logical translation of the statement “some of my friends are not perfect” is ______ $\forall _x (f(x) \wedge \neg p(x))$ $\exists _x (f(x) \wedge \neg p(x))$ $\neg (f(x) \wedge \neg p(x))$ $\exists _x (\neg f(x) \wedge \neg p(x))$
asked Nov 20, 2020 in Discrete Mathematics jothee 204 views
0 votes
1 answer
6
UGCNET-Oct2020-II: 38
What kind of clauses are available in conjunctive normal form? Disjunction of literals Disjunction of variables Conjunction of literals Conjunction of variables
What kind of clauses are available in conjunctive normal form? Disjunction of literals Disjunction of variables Conjunction of literals Conjunction of variables
asked Nov 20, 2020 in Discrete Mathematics jothee 250 views
1 vote
2 answers
7
NIELIT 2016 MAR Scientist C - Section C: 25
Which of the following is FALSE? $Read\ \wedge as\ AND, \vee\ as\ OR, \sim as\ NOT, \rightarrow$ as one way implication and $\leftrightarrow$ as two way implication? $((x\rightarrow y)\wedge x)\rightarrow y$ $((\sim x\rightarrow y)\wedge (\sim x\wedge \sim y))\rightarrow x$ $(x\rightarrow (x\vee y))$ $((x\vee y)\leftrightarrow (\sim x\vee \sim y))$
Which of the following is FALSE? $Read\ \wedge as\ AND, \vee\ as\ OR, \sim as\ NOT, \rightarrow$ as one way implication and $\leftrightarrow$ as two way implication? $((x\rightarrow y)\wedge x)\rightarrow y$ $((\sim x\rightarrow y)\wedge (\sim x\wedge \sim y))\rightarrow x$ $(x\rightarrow (x\vee y))$ $((x\vee y)\leftrightarrow (\sim x\vee \sim y))$
asked Apr 2, 2020 in Mathematical Logic Lakshman Patel RJIT 158 views
0 votes
4 answers
8
NIELIT 2016 MAR Scientist C - Section C: 65
In propositional logic, which of the following is equivalent to $p \rightarrow q$? $\sim p\rightarrow q$ $ \sim p \vee q$ $ \sim p \vee \sim q$ $p\rightarrow \sim q$
In propositional logic, which of the following is equivalent to $p \rightarrow q$? $\sim p\rightarrow q$ $ \sim p \vee q$ $ \sim p \vee \sim q$ $p\rightarrow \sim q$
asked Apr 2, 2020 in Mathematical Logic Lakshman Patel RJIT 531 views
1 vote
2 answers
9
NIELIT 2017 July Scientist B (IT) - Section B: 13
Which of the following statements is false? $(P\land Q)\lor(\sim P\land Q)\lor(P \land \sim Q)$ is equal to $\sim Q\land \sim P$ $(P\land Q)\lor(\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor P$ ... $(P\land Q)\lor(\sim P\land Q)\lor (P \land \sim Q)$ is equal to $P\lor (Q\land \sim P)$
Which of the following statements is false? $(P\land Q)\lor(\sim P\land Q)\lor(P \land \sim Q)$ is equal to $\sim Q\land \sim P$ $(P\land Q)\lor(\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor P$ $(P\wedge Q)\lor (\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor (P\wedge \sim Q)$ $(P\land Q)\lor(\sim P\land Q)\lor (P \land \sim Q)$ is equal to $P\lor (Q\land \sim P)$
asked Mar 30, 2020 in Mathematical Logic Lakshman Patel RJIT 224 views
0 votes
3 answers
10
NIELIT 2017 July Scientist B (CS) - Section B: 16
Which of the following propositions is tautology? $(p\lor q)\to q$ $p\lor (q\to p)$ $p\lor (p\to q)$ Both (B) and (C)
Which of the following propositions is tautology? $(p\lor q)\to q$ $p\lor (q\to p)$ $p\lor (p\to q)$ Both (B) and (C)
asked Mar 30, 2020 in Mathematical Logic Lakshman Patel RJIT 260 views
0 votes
3 answers
11
NIELIT 2017 DEC Scientist B - Section B: 43
Which one is the correct translation of the following statement into mathematical logic? “None of my friends are perfect.” $\neg\:\exists\:x(p(x)\land q(x))$ $\exists\:x(\neg\:p(x)\land q(x))$ $\exists\:x(\neg\:p(x)\land\neg\:q(x))$ $\exists\:x(p(x)\land\neg\:q(x))$
Which one is the correct translation of the following statement into mathematical logic? “None of my friends are perfect.” $\neg\:\exists\:x(p(x)\land q(x))$ $\exists\:x(\neg\:p(x)\land q(x))$ $\exists\:x(\neg\:p(x)\land\neg\:q(x))$ $\exists\:x(p(x)\land\neg\:q(x))$
asked Mar 30, 2020 in Mathematical Logic Lakshman Patel RJIT 445 views
0 votes
0 answers
12
UGCNET-Dec2006-II: 2
The proposition ~ q ∨ p is equivalent to :
The proposition ~ q ∨ p is equivalent to :
asked Mar 27, 2020 in Mathematical Logic jothee 164 views
0 votes
1 answer
13
UGCNET-Jan2017-III: 59
Which of the following statements is true? The sentence $S$ is a logical consequence of $S_{1},\dots,S_{n}$ if and only if $S_{1}\wedge S_{2} \wedge \dots \wedge S_{n}\rightarrow S$ is satisfiable. The sentence $S$ ... of $S_{1},\dots,S_{n}$ if and only if $S_{1}\wedge S_{2}\wedge \dots \wedge S_{n}\wedge S$ is inconsistent.
Which of the following statements is true? The sentence $S$ is a logical consequence of $S_{1},\dots,S_{n}$ if and only if $S_{1}\wedge S_{2} \wedge \dots \wedge S_{n}\rightarrow S$ is satisfiable. The sentence $S$ is a logical consequence of $S_{1},\dots,S_{n}$ if ... logical consequence of $S_{1},\dots,S_{n}$ if and only if $S_{1}\wedge S_{2}\wedge \dots \wedge S_{n}\wedge S$ is inconsistent.
asked Mar 24, 2020 in Discrete Mathematics jothee 681 views
0 votes
3 answers
14
UGCNET-Jan2017-II: 2
Match the following : ...
Match the following : ...
asked Mar 24, 2020 in Mathematical Logic jothee 243 views
13 votes
5 answers
15
GATE CSE 2020 | Question: 39
Which one of the following predicate formulae is NOT logically valid? Note that $W$ is a predicate formula without any free occurrence of $x$. $\forall x (p(x) \vee W) \equiv \forall x \: ( px) \vee W$ ... $\exists x(p(x) \rightarrow W) \equiv \forall x \: p(x) \rightarrow W$
Which one of the following predicate formulae is NOT logically valid? Note that $W$ is a predicate formula without any free occurrence of $x$. $\forall x (p(x) \vee W) \equiv \forall x \: ( px) \vee W$ $\exists x(p(x) \wedge W) \equiv \exists x \: p(x) \wedge W$ ... $\exists x(p(x) \rightarrow W) \equiv \forall x \: p(x) \rightarrow W$
asked Feb 12, 2020 in Mathematical Logic Arjun 5.9k views
5 votes
3 answers
16
ISRO2020-73
Given that $B(a)$ means “$a$ is a bear” $F(a)$ means “$a$ is a fish” and $E(a,b)$ means “$a $ eats $b$” Then what is the best meaning of $\forall x [F(x) \to \forall y(E(y,x)\rightarrow b(y))]$ Every fish is eaten by some bear Bears eat only fish Every bear eats fish Only bears eat fish
Given that $B(a)$ means “$a$ is a bear” $F(a)$ means “$a$ is a fish” and $E(a,b)$ means “$a $ eats $b$” Then what is the best meaning of $\forall x [F(x) \to \forall y(E(y,x)\rightarrow b(y))]$ Every fish is eaten by some bear Bears eat only fish Every bear eats fish Only bears eat fish
asked Jan 13, 2020 in Mathematical Logic Satbir 953 views
1 vote
2 answers
17
CMI2019-B-5
In the land of Twitter, there are two kinds of people: knights (also called outragers), who always tell the truth, and knaves (also called trolls), who always lie. It so happened that a person with handle @anand tweeted something offensive. It was not known ... $3:$ My lawyer always tells the truth. Which of the above suspects are innocent, and which are guilty? Explain your reasoning.
In the land of Twitter, there are two kinds of people: knights (also called outragers), who always tell the truth, and knaves (also called trolls), who always lie. It so happened that a person with handle @anand tweeted something offensive. It was not known ... Suspect $3:$ My lawyer always tells the truth. Which of the above suspects are innocent, and which are guilty? Explain your reasoning.
asked Sep 13, 2019 in Mathematical Logic gatecse 244 views
3 votes
1 answer
18
Mathematical Logic Ques:Self doubt
“Not every satisfiable logic is valid” Representation of it will be $1)\sim \left ( \forall S(x)\rightarrow V(x) \right )$ or $2)\sim \left ( \forall S(x)\vee V(x) \right )$ Among $1)$ and $2)$, which one is correct? and why?
“Not every satisfiable logic is valid” Representation of it will be $1)\sim \left ( \forall S(x)\rightarrow V(x) \right )$ or $2)\sim \left ( \forall S(x)\vee V(x) \right )$ Among $1)$ and $2)$, which one is correct? and why?
asked Jun 4, 2019 in Mathematical Logic srestha 415 views
2 votes
1 answer
19
Doubt on GATE Question
Read the statements: All women are entrepreneurs. Some women are doctors. Which of the following conclusions can be logically inferred from the above statements? All women are doctors All doctors are entrepreneurs All entrepreneurs are women Some entrepreneurs are doctors ... Is it because , if we make set of doctor as 0, then All doctors are entrepreneurs is meaningless.
Read the statements: All women are entrepreneurs. Some women are doctors. Which of the following conclusions can be logically inferred from the above statements? All women are doctors All doctors are entrepreneurs All entrepreneurs are women Some entrepreneurs are doctors Why here $2)$ ... ans?? Is it because , if we make set of doctor as 0, then All doctors are entrepreneurs is meaningless.
asked Jun 1, 2019 in Mathematical Logic srestha 198 views
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