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[ Edited] My question : how to formulate the principle of non-contradiction in order to show clearly that the proposition " all unicorns are both mammals and non-mammals" is compatible with it?

Does this formulation meet the condition just stated : 

       For all object x, and all predicate P, ~ [ P(x) & ~ P(x) ]   ?

Remark : this is not a sceptical question against set theory.

Explanation :

One learns in basic set theory that :

The null set is a subset of any arbitrary set.

From this one can infer, demonstratively, that the following sentences are both true :

(1) All unicorns are mammals. ( The set of unicorns is a subset of the set of mammals).

(2) All unicorns are non-mammals. ( The set of unicorns is a subset of the complement of the set of mammals).

It's clear that both sentences can be false at the same time, but how can they be both true?

The difficulty I see in understanding the phenomenon not in accepting it ( for,as I said, the conjunction is proven in set theory) is : (1) implies there is no x such that x is a unicorn and x is not a mammal (= a non-mammal) ; and (2) says that for all x, if x is a unicorn, x is a mammal.

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    They are both true because there are no unicorns. Thus $\text {Uni}(x) \to \text {Whatever}(x)$ is true for every $x$ (and every predicate whatever). – Mauro ALLEGRANZA Apr 01 '19 at 14:45
  • And the fact that both are true does not contradict Non Contradiction, because the two are not one the negation of the other. – Mauro ALLEGRANZA Apr 01 '19 at 15:01
  • @ Mauro Allegranza - Clearly, it is easy to prove that both are true. But how to formulate the principle of non-contradiction in order to show that " Unicorns are all both mammals and non-mammals" is compatible with this principle? –  Apr 01 '19 at 15:10
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    From $\forall x (\text {Uni}(x) \to \text {Mam}(x))$ and $\forall x (\text {Uni}(x) \to \lnot \text {Mam}(x))$ we have - by FOL - $\forall x (\text {Uni}(x) \to (\text {Mam}(x) \land \lnot \text {Mam}(x)))$. – Mauro ALLEGRANZA Apr 01 '19 at 15:15
  • Again, the conclusion is true because $\lnot \exists x \text {Uni}(x)$. – Mauro ALLEGRANZA Apr 01 '19 at 15:15
  • @ Mauro Allegranza - What would you say about the principle of non-contradiction? Is it even possible to formulate it in predicate logic? Wouldn't we need to quantify over predicates? –  Apr 01 '19 at 15:18
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    In FOL it will be a "schema" : $\forall x \lnot (A(x) \land \lnot A(x))$. – Mauro ALLEGRANZA Apr 01 '19 at 15:22
  • @Mauro Allegranza. Thanks. –  Apr 01 '19 at 15:23
  • But we have also the "general" form $\lnot (\varphi \land \lnot \varphi)$. – Mauro ALLEGRANZA Apr 01 '19 at 15:24

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Your difficulty comes from the fact that you treat this inclusion using natural language, which is to say that when we talk about an implication, then it is not a vacuous one.

When you say "If it rains, I will wear a coat" and you're not wearing a coat, you'd likely receive the answer "But it's not raining..." or something similar.

When you say that "All the people who voted for Asaf Karagila in the US 2016 Presidential Elections are from Mars", you are likely to get a weird look since I am not a US citizen, and who the hell are you talking about anyway?

But this is why it is important to discard natural language and stick to the definitions. When we say that $\varnothing\subseteq X$ we only give a conditional implication, we are not promising that this condition is at any point true and the implication is meaningful.

So when you say that all unicorns are mammals and also not mammals, you're really saying that there are no unicorns. Yes, from a natural language point of view, it's a really odd way of phrasing this fact, but mathematically speaking, this is not a real issue.

Asaf Karagila
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  • @ Asaf Karagila. You say that " the null set is included in any arbitrary set X" is a CONDITIONAL implication. An implication has the form : (X --> Y). A CONDITIONAL implication has the form : X --> ( Y--> Z). According to you, is the sentence : "For all x, if x belongs to the null set, then X belongs to X" a conditional implication? Has it got the first form or the second? –  Apr 03 '19 at 21:07
  • Don't get hung up on terms here. I might be a set theorist, but I am not a philosopher, and philosophy is taking logic more in its syntactic form. I may have misused "conditional implication". When I said conditional implication, I meant that we are not making any promises about the premise of the implication being true. – Asaf Karagila Apr 03 '19 at 21:09
  • @ Asaf Karagila.Thanks. Coming back to the main point, could you - when you have some free time - throw some light on the way the antique principle of non contradiction ( or law of non contradiction) could be stated from a set theoretic point of view. As a philosopher, it is always a great pleasure for me to see how set theory gives new insights in old topics ( be these topics logical or ontological as in Mario Bunge's books). –  Apr 03 '19 at 21:19