# Is the argument "a is b" but "b is not a" valid?

Is it true in logic that "a is b" but "b is not a"? Does it work one way or both ways?

For example:

``````a is b
Therefore, b is not a
``````

If this is valid, how can you prove that it is valid?

• This looks similar to the most common logical fallacy I see people make. Namely: If A implies B, does B imply A? And the answer is "No". But this doesn't mean "If B then Not A" - but rather "If B then "Not Necessarily" A" - ie we can't tell the value of A if B is True. It's not the same, but may be the source of your thoughts.
– Ryno
Oct 5 '12 at 8:07
• Entirely depends on the relationship between A and B.
– user2411
Oct 5 '12 at 11:17
• Water is blue, blue isn't water Oct 7 '12 at 7:28
• @ViniyoShouta: "Water is Blue" means "Water has the property of Blueness"; it is not equivalence. Oct 7 '12 at 15:28
• There surely is a typo in this question? Do you mean "a is b, therefore, b is a"? As currently expressed, under whatever interpretation, it is clearly false. Oct 10 '12 at 14:31

It was an old anecdote "It depends on a definition what is is", but it seems to answer your question.

There are several formal meanings to is:

1. `A is B` stands for "any object `x` that belongs to a category `A`, also always belongs to a category `B`". In this case, `A` is a subset of `B`. One may say, any odd number is a number, and it's valid. Obviously, the opposite is not valid: any number is odd number;
2. `A is B` stands for "objects `A` and `B` are equal". Likewise "My Cat and Cat Living In My House are the same terms". Equality is symmetric;
3. Similar to above, but `A is B` applies to categories: if category `A` equals to category `B`, that means that any object `x` belonging to `A` also belongs to `B` and, vice versa, any `y` belonging to `B`, also must belong to `A`.

So, answering your question, variant (1) perfectly allows `A is B` and `B is not A`

• I'm looking for what it means in syllogistic logic. Oct 6 '12 at 0:26
• In syllogistic logic you need at least two premises to derive the conclusion. In your example there is only one "A is B" is the first. It may be minor premise, it may be major premise. Oct 12 '12 at 22:29

In classical logic, equivalence is symmetrical. If A is B, then B is A.

• I'm looking for what it means in syllogistic logic. Oct 6 '12 at 0:26
• The same. I don't know of any form of syllogistic logic where equivalence is not symmetrical. Oct 7 '12 at 15:26

While you've already gotten a few answers, it seems like you are having trouble extrapolating from classical logic or predicate logic to syllogistic logic.

It's easy to understand how categorical propositions interact in syllogistic logic, when you understand the square of opposition and the basic ontology of categorical propositions.
Every categorical proposition (e.g. "a is b"), can be reduced to one of these four logical forms. It is a true statement that every proposition can only satisfy one of the forms. It is a false statement that for every subject that is a predicate, that predicate is not the subject. To see a proof of this falsehood, consider the statement "A dog is a canine". If we let a = "a dog", and let b = "a canine", then by your proposed reasoning, a dog is a canine, but a canine is not a dog. Since the terms dog and canine refer to the same thing, this can't be the case.

Take a moment to examine the following diagram: Each of the connecting lines represent a relationship between propositions about the same term. These relationships are explored in depth in the articles I linked to earlier. However, I want to explain where I think you may have gotten confused. It seems like you are trying to apply rules about relations between propositions as if they were rules about relations between entities. One type of relation between propositions is called "contradictory". It means there is no case when the two propositions are true. An entity cannot be contradictory. It doesn't make sense. Back to the earlier example, a dog cannot be contradictory. However, statements about a dog (or dogs) can be contradictory (e.g. "All dogs are furry" [SaP] and "I shaved my dog" [SoP]).

Now, from some proposition forms we can extrapolate the negation of another proposition forms. For example, if we know that "All dogs are canines" [SaP], we also know that it is not the case that "Some dogs are not canines" [SoP]. However, this relationship is not commutative. If we know that "some dogs are not furry" [SoP], we cannot know whether or not "all dogs are not furry" [SeP]. There just isn't enough information (notice how the arrow is only on one side of the line).

My guess is that the misconception arose during the classification of entities that satisfy a predicate, such as "furry dogs" standing in for all entities such that the entity is a dog and the entity is furry. Then we end up with an instance of a class and a class, and we compare them using the terminology that you used. So we get a = "Rover is a furry dog", but it is false that "a furry dog is Rover", because Rover is a concrete instance of a "furry dog" and "a furry dog" is an abstract instance - a universal. A furry dog is equal to the class of entities that are dogs and are furry, but a Rover is simply a particular entity that is a dog and is furry.

The Sameness In Reality:

1) Something has the sameness to something else,

• then both at different places and separated each other, OR

2) Something has the sameness to something itself,

• then both must be pointed at the same place as the one thing.

Again, In Reality

• A is B and B is A, must be understood as "If something has the sameness to something else, then both must be pointed at the same place as the one thing (otherwise B is not A, THERE IS LITTLE BIT DIFFERENCE, WHATEVER IT IS)"

• A is B and B is not A, must be understood as "If something has the sameness to something else, then both at different places and separated each other (otherwise B is A, AS ONE, THERE IS NO DIFFERENCE AT ALL)".

Meaning if we found the sameness in between the two things, then at least both have difference, both are pointed at different places.

Here is a natural deduction proof showing that if a = b then b = a. If a = b also implied not b = a then first order logic with identity would be inconsistent: Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

• That's assuming "a is b" uses the "is" of identity and not the "is" of predication. Aug 20 '19 at 4:05