2,631 views
0 votes
0 votes

If the propostion 'All thieves are poor' is false, which of the following propositions can be claimed certainly to be true?

  1. Some thieves are poor
  2. Some thieves are not poor
  3. No thief is poor
  4. No poor person is a thief

3 Answers

Best answer
3 votes
3 votes

It can be easily proved.

Let $T(x)$ denotes $x$ is a thief and $P(x)$ denotes $x$ is poor, then All thieves are poor is  $\forall x(P(x) \to T(x))$

Since given statement is false, take the negation

$\neg ( \forall x(P(x) \to T(x))) =  \exists x(\neg(P(x) \to T(x))$          // Rewrite $A \to B$ as $A' \vee B$

$= \exists x(\neg (P(x)'  \vee T(x)))$ 

$=\exists x(P(x) \wedge T(x)')$

Which means " there are some thieves who are not poor"

option 2

selected by
2 votes
2 votes

CASE1: "ALL theives are poor" is false when atleast there is one/some  theif/theives  who is/are not poor"

CASE2: "ALL theives are poor" is false "NO theives are poor"

but we cannot be sure that due to which case the "ALL theives are poor" is false.

but one thing is sure if "ALL theives are poor" is false...then there must be some theives who are not poor.

if domain consist of all the theives

p(x) = "x is poor"

 ~∀x(p(x))  =  ∃ x(~p(x))

....option 2 is correct

edited by
0 votes
0 votes
thieves are poor = p(x)

given; for all p(x) =false

~{for all p(x)} =true

some ~p(x) = true

'some thieves are not poor' Ans(2)

Related questions

2 votes
2 votes
2 answers
4
shivani2010 asked Jun 12, 2016
4,252 views
An undirected graph is Eulerian if and only if all vertices of G are of the sum of the degrees of all nodes isA. Same degreeB. ODD degreeC. Need not be ODDD. ...