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I saw an example in a book talking about this setup:

All 'P' are 'Q'

All 'R' are 'P'


All 'R' are 'Q'

I am wondering if there are any real-life examples where the premises are true, but the conclusion comes up false.

In trying examples, either a premise was made false by the other premise or conclusion, or the premises resulted in a true conclusion

Otherwise, does every real life example fit the formula, meaning the logic is unbreakable?

Example setup:

All monkeys are sapiens

All humans are monkeys


All humans are sapiens. -----How to get the conclusion false?

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This is one of the classic 24 valid syllogisms, which means: It's a correct logical argument. In first-order logic, the premises can be written as ∀x(P(x)→Q(x)) and ∀x(R(x)→P(x)), and this implies ∀x(R(x)→Q(x)). So, whenever the premises are true, then the conclusion is also true.

Except if you cheat.

What does "cheat" mean? Well, for instance, words in a natural language may have several meanings, and the meaning may implicitly depend on the context. So you may agree that "all stars are celestial bodies" and "all Grammy winners are stars", but since the word "stars" has a different meaning in the first and in the second sentence, you should better not deduce "all Grammy winners are celestial bodies".

Additionally, statements in a natural language may hold under certain implicit additional conditions. "All Greeks are citizens of the European Union" may be uncontroversial, and "all pupils of Socrates are Greeks" as well, but the first statement comes with an implicit "now" and the second with an implicit "2400 years ago", so you should not deduce "all pupils of Socrates are citizens of the European Union". Similar problems may arise, if the first statement holds legally, but not in practice, and the second one holds in practice, but not legally. Then the conclusion may hold neither legally nor in practice.

The problem is that in these cases the translation of the natural language sentences into logical formulas of the form ∀x(P(x)→Q(x)) and ∀x(R(x)→P(x)) is inadequate. Superficially, the examples seem to break the rule, but it's just abuse of linguistic ambiguity.

  • Did you mean "this implies ∀x(R(x)→Q(x))"? – cHao Aug 25 '17 at 13:14
  • @cHao Yes, of course. Thanks for spotting the typo. – Uwe Aug 25 '17 at 13:16
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This sort of First Order Logic is "sound" as proven in the Soundness and Completeness Theorems of First Order Logic constructed by Kurt Gödel in the early 20th century. That means that for any logically valid deduction in this system, if the premises are true, then the conclusion must also be true (because the logical system is sound).

With this in mind, you can construct many different interpretations of various logical deductions (some with true premises and some without). Presumably, if you construct an interpretation that contains factual premises, then those premises will be true and, therefore, so will the deduction's conclusion. Thus, there should not exist any "real world" examples that "break" this logical system unless they rely on some form of linguistic equivocation.

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This logical law called syllogism can be formulated, using set theory, such that its truth immediately becomes obvious: Let A be a subset of B and B be a subset of C, then A is a subset of C too. A prime number is a natural number, a natural number is an integer, so a prime number is an integer.

You cannot break this law other than by cheating, i.e., applying two different meanings of identity, for instance in a temporal development: Every tadpole is a frog, every frog can croack, so every tadpole can croak.

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How to get the conclusion false?

I regret to write that you are stuck with a valid syllogism. This example is AAA in the first figure.

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An interesting fact is that quantum mechanics can break logic. An example for this is Bell's inequality.

Let's take 3 properties: A,B,C. Now it is easy to see that for a set of everyday objects the number of objects that have properties A and not B plus the number of objects that have properties B and not C is greater than number of objects with properties A and not C. This is very basic math.

In formula: #(A and not B)+ #(B and not C)>= #(A and not C).

In fact by this equation (more precisely by the proof behind the equation) we get:

All R are P

All P are Q

All R are Q

Where R=P=an object satisfying A and not C

And Q=an object satisfying A and not B or B and not C.

However for electrons that is not true. There are A,B and C properties of electrons for which: All R are P All P are Q But not all R are Q.

A,B and C are the positivity of angular momentum around axis x,y and z respectively.

The problem is that electrons are not everyday objects. However electrons broke logic and made a valid syllogism invalid. Maybe this is the everyday example you were looking for?

  • Is there a typo anywhere? You have defined R and P identically, so "all P are Q" should be equivalent to "all R are Q". – Uwe Aug 29 '17 at 7:39
  • No, this is not a typo. That was the easiest way for me to put the contradiction into syllogistic form. – MetaLogicianWannabe Aug 29 '17 at 7:48
  • @Uwe this can be easily modified into a syllogism where the predicates are different. – MetaLogicianWannabe Aug 29 '17 at 8:03

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