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While going over my lecture notes for the preparation of exams, I stumbled upon this;

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Everything is making sense except the reflexive and anti symmetric relation of natural numbers.

1) By the definition of inequality , we can say that that $(x, x)$ would not appear as an ordered pair.

2)How can it be anti symmetric if the hypothesis $(x,y) \wedge (y, x)$ does not hold? Wouldn't it be meaningless to say anything about the implication if the hypothesis is false?

Tom Lynd
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2 Answers2

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  1. You're correct. Only the first two relations are reflexive; not the third.

  2. All three relations (even the third) are indeed anti-symmetric. The third one is vacuously anti-symmetric, since as you pointed out the hypothesis is always false. False hypotheses are a bit unintuitive, but they do happen fairly often in logic. As a convention, we interpret implications with false hypotheses to be true.

Adriano
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  • What is the motivation for assigning it a truth value when the hypothesis is false? Why should I not call it false? – Tom Lynd Sep 22 '15 at 16:15
  • Think of a true implication as a promise that has not yet been violated. Imagine a politician who claims: "If you elect me, then I will give everyone a pony." The only way to demonstrate that this implication if false is if we elect him and he doesn't give everyone a pony. If we don't elect him, then his promise has not yet been violated, so on good faith we decide that he is not guilty of lying to us (innocent until proven guilty). – Adriano Sep 22 '15 at 16:22
  • Well, if the promise is not broken, neither is it kept – Tom Lynd Sep 22 '15 at 16:25
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The last relation is, indeed, not reflexive, as you have surmised. However, it is antisymmetric, since there are no $x,y$ such that $$(x\:R\:y)\wedge(y\:R\:x)\wedge(x\ne y).$$ After all, as you said, there are no $x,y$ such that $x\:R\:y$ and $y\:R\:x$ in the first place! So, one might say that it is vacuously antisymmetric.

Cameron Buie
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