Some houses are bricks is surely true but All houses are bricks may be true or false

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+1 vote

If the proposition ‘Houses are not bricks’ is taken to be False then which of the following propositions can be True?

- All houses are bricks
- No house is brick
- Some houses are bricks
- Some houses are not bricks

Select the correct answer from the options given below:

- b and c
- a and d
- b only
- c only

+1 vote

Let $x$ : An object

$h(x)$: $x$ is a house and

$b(x)$: $x$ is a brick

Houses are not bricks $\equiv$ For all $x$ , if $x$ is a house then it is not a brick $\equiv \forall x( h(x) \rightarrow \sim b(x)) \equiv \forall x (\sim h(x)\ \vee \sim b(x))$

This above proposition is considered **false** so will negate it.

$\sim( \forall x (\sim h(x)\ \vee \sim b(x))) \equiv \exists x ( h(x)\ \wedge b(x)) $

- All houses are bricks $\equiv$ For all $x$ , if $x$ is a house then it is also a brick $\equiv \forall x ( h(x) \rightarrow b(x)) \equiv \forall x (\sim h(x)\ \vee b(x))$
- No house is a brick $\equiv$ There does not exist an $x$ such that $x$ is a house and it is a brick $\equiv \sim \exists x ( h(x) \wedge b(x)) \equiv \forall x (\sim h(x)\ \vee \sim b(x) )$
- Some houses are bricks$\equiv$ There exists some $x$ such that $x$ is a house and it is a brick $\equiv \exists x(h(x) \wedge b(x))$
- Some houses are not bricks$\equiv$ There exists some $x$ such that $x$ is a house and it is not a brick $\equiv \exists x(h(x) \wedge \sim b(x))$

$\therefore$ Option $D.$ is the correct answer.

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