Consider the proposition: “If someone in your class has a dog, then everyone in your class has a cat.” -Translate this sentence into mathematics, letting D(x) be the predicate “x has a dog”, C(x) be the predicate “x has a cat.” Let the universe of discourse be the set of students in your class. ~=not, /\=and,/=or, ->=arrow

  • negate:( ∃x)D(x) -> (∀x)C(x)
  • ~( ∃x)D(x) \/(∀x)C(x)
  • ( ∃x)D(x) /\ (∃x)~C(x)
  • translated back into english:

some student have a dog and some students do not have a cat

  • $\begingroup$ I'm confused. Why have you written 3 logical sentences? Do you want us to pick which one is equivalent to the English you started with? $\endgroup$ – Nathaniel Mayer Jan 30 at 17:41
  • $\begingroup$ I translated it into a math problem then negate it, then I translated it back into an English statement although translating back into english is not important, just seeing if i understand this.@NathanielMayer $\endgroup$ – Derek Long Jan 30 at 17:43
  • $\begingroup$ So what's the second sentence? Is it supposed to be equivalent to the first? (It's not) $\endgroup$ – Nathaniel Mayer Jan 30 at 17:46
  • $\begingroup$ yes I was trying to make it equivalent. is ( ∃x)D(x) /\ (∃x)~C(x not correct?@NathanielMayer $\endgroup$ – Derek Long Jan 30 at 17:49
  • $\begingroup$ The third one is correct as a negation of the first. It's not a negation of the second $\endgroup$ – Nathaniel Mayer Jan 30 at 17:50

The first sentence you wrote is an accurate translation, and the third sentence is the correct negation of the first.

The second is different from both, not sure what you were doing there. I suspect maybe you were looking for $\vee$ instead of $\wedge$.

  • $\begingroup$ yes I forgot to switch that! Thank you again $\endgroup$ – Derek Long Jan 30 at 17:56
  • $\begingroup$ You can accept the answer :) $\endgroup$ – Nathaniel Mayer Jan 30 at 17:57

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