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Statement : If it is raining, then the home team wins.

Converse : If the home team wins, then it is raining.

Why are these two not logically equivalent? The statement says that if it rains then the home team wins. In converse, since the home team wins, it should rain. Right?

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Is this perhaps confused with the contrapositive of a statement? en.wikipedia.org/wiki/Contraposition – Irregular User Mar 6 at 6:23
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What if it rains and the home team loses. The converse allows that. The statement does not. What if it's sunny and the home team wins. The converse doesn't allow that but the statement does. – fleablood Mar 6 at 6:32
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It's easier to see that, in general, a statement isn't equivalent to its converse if you can find one with a bit more "directionality." For example, $\rm man \implies human$ and yet it is certainly not true that $\rm human \implies man$, as human females exist. (Sorry for the repost: tried to sort out negating \implies with MathJax and the previous slash was driving me nuts) – pjs36 Mar 6 at 7:08
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"if I have 100k math.se reputation then I am a math.se user" is a true statement. Why is it not equivalent to "if I am a math.se user then I have 100k math.se reputation"? (math.se being this site) – immibis Mar 6 at 10:09
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Note that natural language does not always conform to the same conventions as mathematical language. Depending on the surrounding context, the sentence may be interpreted as "the home team wins if and only if it is raining", in which case the converse is equivalent. – Peter Olson Mar 7 at 3:30
up vote 19 down vote accepted

It may help to ask yourself what would have to happen for each of the statements to be false.

(A) If it is raining, then the home team wins.

The only way this can be false is for it to be raining, and the home team loses. (For the same of simplicity I’ll assume that there are no ties.)

(B) If the home team wins, then it is raining.

The only way this can be false is for the home team to win while it is not raining.

Those are two completely different events:

  1. It’s raining, and the home team loses.
  2. It’s not raining, and the team wins.

If the two statements were equivalent, exactly the same circumstances would render them both false. As you can see, that’s not the case, so they cannot be equivalent.

In fact, there are four possible states of affairs:

  1. It’s raining, and the home team loses.
  2. It’s not raining, and the team wins.
  3. It’s raining, and the home team wins.
  4. It’s not raining, and the home team loses.

As we’ve seen, statement (A) is inconsistent with state (1); what about the other three states? It’s obviously consistent with state (3). It’s true but less obvious that it’s also consistent with states (2) and (4): this is because statement (A) says nothing at all about who wins if it’s not raining, so no outcome of the game can disprove it in that case. Thus, statement (A) boils down simply to saying that state (1) isn’t the case: either it’s not raining, in which case (A) says nothing about the outcome of the game, or it is raining, and the home team wins.

I’ll leave it to you to apply similar reasoning to see that statement (B) is consistent with each of the states except state (2).

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Alternatively, you can show logical equivalence with a truth table. Let $r = \{\text{Rains}\}, w = \{\text{Home team wins}\}$. Then

\begin{array}{c|c|c|c} r&w&r\to w&w\to r\\\hline T&T&T&T\\\hline T&F&F&T\\\hline F&T&T&F\\\hline F&F&T&T \end{array} Notice that the conditional statements $r\to w$ and $w\to r$ have different truth tables, and hence, they are not logically equivalent.

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The original statement says that if it's raining, the home team wins. It is also possible, based on this statement, that even if it's not raining, the home team still wins.

The converse statement says that if the home team wins, then it's raining. It is also possible for it to rain without the home team winning.

However: Under this converse, it is not possible for the home team to win without it raining. Conversely (!), under the original statement, it is not possible for it to be raining without the home team winning.

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The statement only says that if it is raining then that's enough to know that the home team wins.

So raining guarantees home win, but it does not say that this is the only condition under which the home team wins.

It could be true that when it snows the home team also wins. So you can see that the fact that the home team wins does not guarantee anything about rain.

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Yes. The statement does not say: Only if it is raining will the home team win. – Dan Christensen Mar 6 at 16:31

An intuitive example:

If I have more than 10 coins, I have more than 5 coins.

If I have more than 5 coins, I have more than 10 coins.

You wouldn't regard the above as equivalent, would you?

The first sentence is always true, no matter what. The second can be false, for instance when you have 7 coins.

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Consider the following:

If you have studied, then you will pass the test.

Let's say you accept that it is true. Still, you might believe that some people don't have to study but will pass the test anyway! Such people would provide counter-examples to the following:

If you will pass the test, then you have studied.

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Geometrically, you can think of implication ($\implies$) as inclusion ($\subseteq$). If we're working in some first-order structure $M$, such that $$M \models \forall x \left( \varphi(x) \rightarrow \psi(x) \right),$$ ("$\models$" is to be read as "satisfies") then for any $a \in M$, if $M \models \varphi(a)$, then $M \models \psi(a)$.

This means: the definable set cut out by $\varphi(a)$ $$\varphi(M) \overset{\operatorname{df}}{=} \{a \in M \operatorname{ | } M \models \varphi(a)\}$$ is contained in the definable set $\psi(M)$ cut out by $\psi(a)$. That is, $$M \models \forall x \left(\varphi(x) \rightarrow \psi(x)\right) \iff \varphi(M) \subseteq \psi(M).$$

(For example, let $\varphi(x)$ be the formula in the language of rings which says that $x$ is a root of a polynomial $p(t)$, and let $\psi(x)$ say that $x$ is a root of $p(t)q(t)$ for some other polynomial $q(t)$.)

To answer the question in your title, if all first-order implications were equivalent to their converses, then there would be no proper inclusions of definable sets in any first-order structure $M$. Which is not the case---in any nonempty structure, the empty set (falsehood) is always properly contained in (hence implies) the entire structure (truth).

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consider p:it is raining and q:home team wins so the compound statement is if p then q.that is p->q and converse is q->p.so by identities p->q = ~p V q and q->p = ~q V p = p V ~q

so ~p V q is never equivalent to p V ~q thus the statement is not equivalent to its converse.

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In this type of statement, causation works in one direction. The result does not do the causation. Perhaps the raining/wins combination is not truely an example of a "statement" and its "converse."

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