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I have a problem in proving invalid the following argument:

Horses and cows are mammals.Some animals are mammals.Some animals are not mammals.Therefore all horses are animals.

If we translate it into the logical notation then we have :

The premises are :

$(\forall x)(Hx \lor Cx \rightarrow Mx)$.

$(\exists x)(Ax \land Mx)$

$(\exists x)(Ax \land \sim Mx)$

The conclusion is :

$(\forall x)(Hx \rightarrow Ax)$.

Where $Ax$ is means $x$ is an animal, $Hx$ means $x$ is a horse, $Cx$ means $x$ is a cow, $Mx$ means $x$ is a Mammal.

I have to show it is an invalid argument.But the 2nd and 3rd premises are like contradictory to the conclusion part.

As per I know if I can show a truthvalue assignment for which the conclusion is false still the premises are true,then the argument will be invalid. But I am unable to find such a truthvalue assignment, looking for a help. Thanks.

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  • $\begingroup$ From your premises, do you know that all mammals are animals? :-) $\endgroup$ – Francesco Polizzi Aug 21 '17 at 15:44
  • $\begingroup$ Yes, it helped me to get the flaw. $\endgroup$ – hiren_garai Aug 21 '17 at 16:19
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Red cars and red trucks are red. Some baloons are red. Some baloons are not red. Therefore, all red cars are baloons.

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You just need to show that it's valid for there to be a horse that is not an animal.

So take $y$ such that $Hy$ is true and $Ay$ is false. Note how this doesn't interfere with the second or third premise.


To get the full truth values, consider a universe $\{x,y,z\}$ such that $$ Mx, My, Ax, Az, Hy\quad \text{are true} $$ and $$ Cx, Cy, Cz, Hx, Hz, Ay, Mz\quad \text{are false}. $$ Then verify that all premises hold and the conclusion fails.

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  • $\begingroup$ Yes, I have taken that, but I can't find the truth values such that the premises are true. $\endgroup$ – hiren_garai Aug 21 '17 at 15:46
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    $\begingroup$ @HirenGarai Well by the first premise you must have $My$ is true. Then you can have $Cy$ be either true or false. This doesn't interfere with the 2nd or 3rd premise because those just guarantee the existence of some $x$ and some $z$ such that $Ax$ and $Mx$ are true and $Az$ is true and $Mz$ is false. So overall you can take your "universe" to be $\{x,z,y\}$ such that $Hy, Ax, Az, Mx, My$ are true and all others are false. $\endgroup$ – John Griffin Aug 21 '17 at 15:55
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    $\begingroup$ I am ok with your answer . $\endgroup$ – hiren_garai Aug 21 '17 at 16:10
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    $\begingroup$ Thanks a lot for the detailed answer sir !!! Now its ok. $\endgroup$ – hiren_garai Aug 21 '17 at 16:21

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