# Proving invalidity of an argument.

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.

• From your premises, do you know that all mammals are animals? :-) – Francesco Polizzi Aug 21 '17 at 15:44
• Yes, it helped me to get the flaw. – hiren_garai Aug 21 '17 at 16:19

Red cars and red trucks are red. Some baloons are red. Some baloons are not red. Therefore, all red cars are baloons.

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.

• Yes, I have taken that, but I can't find the truth values such that the premises are true. – hiren_garai Aug 21 '17 at 15:46
• @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. – John Griffin Aug 21 '17 at 15:55
• I am ok with your answer . – hiren_garai Aug 21 '17 at 16:10
• Thanks a lot for the detailed answer sir !!! Now its ok. – hiren_garai Aug 21 '17 at 16:21