0
$\begingroup$

∃x (P(x) → Q(x)), ∃x P(x) ⊨ ∃x Q(x)
I am trying to find the invalidity of the following sequents.
A = Set of natural number
P(x) : x is odd
Q(x) : x is not divisible by 2

What i don't understand is how can this be invalid ? There will always exist an x that is not divisible by 2 . How can we make this false.
If i assume another model such that
A = {0,1}
P(x) = {0}
Q(x) = {}
Now in this case if we assume q(x) to be an empty set it will not be in the model .. So the invalidity shows .Does that make sense?

$\endgroup$
2
  • $\begingroup$ First of all you should be more specific and improve the mathematical expressions. They are wrong (in my opinion) and only very hard to understand. What is $A$? Is $x \in A$? If you are not able to express your thoughts in mathematical formulas you should try to explain your problem in a few words :-) $\endgroup$ – Elastic Lamb Oct 9 '17 at 19:56
  • $\begingroup$ In your example: why is $P(x) = \{0\}$? $0$ is even and not odd. Or do I not understand you? $\endgroup$ – Elastic Lamb Oct 9 '17 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.