# Anomaly set theoretic equivalent to material implication

1. May 16, 2015

OK, this is embarrassing, but I never looked carefully at this elementary point. We say that if
p implies q
P is the set of all things for which p is true
Q is the set of all things for which q is true
then Q ⊆ P.
Also that the set of all things for which p&q is true equals P∩Q
But p & q implies p, so (from the above) P ⊆ P ∩ Q, which is in general false.
What is wrong?
Thanks

2. May 16, 2015

### micromass

Are you sure that shouldn't be $P\subseteq Q$?

3. May 16, 2015

that was also my reaction, micromass, but I read a number of sources which stated it as I have put it, including HallsofIvy on this forum: https://www.physicsforums.com/threads/set-theory-representation-of-material-implication.206811/ :
/If P is the set of all things for which statement p is true and Q the set of all things for which statement q is true, then "If p then q" can be represented as "Q is a subset of P"./
This is apparently the reason that the symbol for material implication is often the (backward) subset sign: "p implies q" is often written as p⊃q, as you can see in http://en.wikipedia.org/wiki/Material_conditional and other sources. (In fact, this usage is what got me started on this question.)
So .....????

4. May 16, 2015

### micromass

Let p = divisible by 4, let q = divisible by 2. Then $P = 4\mathbb{Z}$ and $Q = 2\mathbb{2}$. We have $p\rightarrow q$ and we have $P\subseteq Q$. So I don't see how it could be differently.

The answer of Halls in that thread is wrong on another level to. It is right that the set theoretic version of $p \wedge q$ is $A\cap B$ and so on. But the set theoretic version of a operation should be a set. So it is false that the set theoretic analog of $p\rightarrow q$ is $A\subseteq B$ (or $B\subseteq A$), since that is not a set. The set theoretic analog is rather equal to $A^c \cup B$ (in classical logic, the theory becomes much more complicated and more beautiful in other logics).

The reason why we sometimes use $p\subset q$ for $q\rightarrow p$ is an unfortunate historical coincidence, see http://math.stackexchange.com/quest...the-symbol-supset-when-it-means-implication-a

5. May 16, 2015