Are all true statements equivalent?

Question

I am wondering whether all true statements are equivalent and if yes, in which sense.

Let A and B be true statements. We derive $A \quad \Longleftrightarrow \quad A \vee B \quad \Longleftrightarrow \quad B$

This would mean

$1 = 1 \quad \Longleftrightarrow \quad (1=1) \vee \pi \text{ is irrational} \quad \Longleftrightarrow \quad \pi \text{ is irrational}$.

However, this does not make any sense to me. By the same token, one could show $1 = 1$ being equivalent to Fermat's last theorem (which we know is true) or any other statement that is known to be true.

Maybe this is indeed correct, or I am mixing up different notions of equivalence.

A or B is true because B is true. We get true and true on both sides of the implication arrow. This evaluates to true. However, I suppose that I am mixing up two different formalisms here. — shuhalo, Dec 17 '17 at 07:53
$A$ implies $A\lor B$. $A$ is not equivalent to $A\lor B$ Take for example $A:=\text{I am asleep}$ and $B:=\text{I am awake}$. — , Dec 17 '17 at 08:02
In your example, if I know that 'I am awake' is true, then 'I am awake or I am asleep' implies 'I am awake'. https://en.wikipedia.org/wiki/Material_conditional#Truth_table — shuhalo, Dec 17 '17 at 08:07
True statements? Truth is relative to a structure. True where? — Asaf Karagila, Dec 17 '17 at 08:11
Ah, you mean you already know that $A$ and $B$ are true? Then, of course $A\Leftrightarrow B$ is true (and so is $A\Leftrightarrow A\lor B$). However, it is pretty vacuous too, it doesn’t help establish the truth where you don’t know it. — , Dec 17 '17 at 08:15
It's a good question. Technically yes. 1=1 if and only if pi is irrational iff FLT is true iff ducks aren't dolphins. One thing important is those statements have no variables which is probably crucial. Fred is a prime number's truth value relies upon the value of what Fred is. In proofs we only care about what is inferable by conditions. Not because the truth of absolute statements aren't important, but because they have no predictive value. — fleablood, Dec 17 '17 at 08:39
One problem you have with your question is that you are not using language precisely. So the logical connectives you are using are binary. But if you put brackets in eg $(A \iff (A\vee B))\iff B$ the statement you get is not true. Note also that if $A$ and $B$ are both equivalent to "true" they are equivalent to each other. — Mark Bennet, Dec 17 '17 at 08:45
Do you understand the answers that Bram28 and I have given you? — user21820, Jan 12 '18 at 16:03

Peter Smith · Answer 1 · 2017-12-23T21:46:42.327

(A) You need to distinguish what is often called material equivalence from logical equivalence

To keep things simple, assume $A$ and $B$ are interpreted wffs, coming with some understood semantics, which fixes their truth-values.

To say that $A$ is materially equivalent to $B$ is just to say that $A \leftrightarrow B$ is true -- i.e. both $A \to B$ and $B \to A$ are true, where $\to$ is the usual truth-functional conditional. So, just by the truth-table, as you say, if $A$ and $B$ are both true on the given semantics, then $A \leftrightarrow B$, $A$ and $B$ are materially equivalent.

To say that $A$ is logically equivalent to $B$ is just to say that $A \dashv\vdash B$ is true -- i.e. both $A \vdash B$ and $B \vdash A$ are true, where $\vdash$ is entailment in some assumed background logic. Plainly, fixing the truth-values of $A$ and $B$ doesn't fix whether the wffs logically entail each other.

(The notions are related: $A$ and $B$ are logically equivalent if and only if it is a logical theorem that they are materially equivalent: so $A \dashv\vdash B$ if and only if $\vdash A \leftrightarrow B$. Which brings out clearly that logical equivalence, the truth of $\vdash A \leftrightarrow B$, is stronger than material equivalence $A \leftrightarrow B$.)

(B) Let's now complicate things a bit in two ways. First, we will want to distinguish semantic entailment in a logic ("on every interpretation ..."), and syntactic entailment ("there is a formal proof from ..."), so that gives us two senses of logical equivalence. Second, we might want to liberalize from the notion of pure logical entailment/equivalence to some notion of entailment/equivalence relative to some background theory.

But the key point remains: mere material equivalence (sameness of truth-value) and logical/theoretic equivalence (inter-derivability perhaps with respect to some background theory) are distinct -- and unqualified talk of "equivalence" can be (potentially) confusing.

(C) Unfortunately, $\Leftrightarrow$ can be used by different authors either way to mean $\leftrightarrow$ or $\dashv\vdash$, either to mean mere material equivalence or logical equivalence. So caution is needed!

Bram28 · Answer 2 · 2017-12-19T17:59:58.113

The short answer is: No.

Two statements being true means that relative to some 'situation', 'scenario', or 'world', both statements are true, while two statements being equivalent means that the two statements are true in the exact same situations.

For example, suppose that today I am wearing my red shirt and it is also raining outside. Then the statements 'I am wearing a red shirt' and 'It is raining outside' are both true today. But they are not equivalent, because obviously it is not always the case that exactly whenever I am wearing a red shirt it is raining outside.

Longer answer:

First of all, by 'equivalence' I am using the notion of logical equivalence (for which we indeed typically use $\Leftrightarrow$), rather than 'material equivalence' (for which we typically use $\leftrightarrow$). You understand the difference, right?

Now, statements like 'I am wearing a red shirt' and 'it is raining' are clearly about contingent matters, for which it is easy to see that both being true is not the same as them being equivalent. But what about mathematics, where 'truth' is typically understood as 'necessary truth'? That is, aren't mathematical statements like '$1=1$' and '$\pi $ is irrational' always true? And, as such, whenever two mathematical statements are true, aren't they true in the exact same scenarios, namely every scenario?

Well, it turns out that even mathematical statements can be seen as contingent, namely contingent upon the axioms we are assuming. The angles of a triangle adding up to the sum of two right angles is only true in Euclidian geometry, for example.

Now, as it so happens, $1=1$ is typically seen as a 'deeper' truth yet, namely as a logical truth, and those are typically regarded as necessary truths: in every world, no matter how we interpret '$1$', it will be identical to itself. But '$\pi$ is irrational' is not seen as a logical truth, but 'only' as a mathematical truth.

So, as you can see, things get a good bit complicated, and philosophers of mathematics are still discussing what 'mathematical truth' really is, so what i just wrote in the longer answer is far from definite. But, I think my short answer should be sufficient for you to realize the basic distinction between two statements being true, and them being equivalent.

"while two statements being equivalent means that the two statements are true in the exact same situations" -- well, in one usage of equivalence. But since the OP was asking whether he was mixing together difference notions of equivalence, it is worth taking up that thought and noting that (rightly or wrongly, more or less commonly) another weaker notion is also in play in some discussions, leading to possible confusion. — Peter Smith, Dec 17 '17 at 14:51
So the short answer is: no, true statements do not imply each other. This is unlike with false statements, which do imply each other, right? Ex Falso Quodlibet. — shuhalo, Dec 17 '17 at 16:32
@shuhalo No, false statements do not imply each other either. The 'Falso' is the contradiction. And yes, a contradiction (i.e a necessarily* false statement) implies anything ... but that is not true for just a plain false statement. I can use the same example. Suppose I am not wearing my red shirt and it is not raining. Then 'I am wearing a red shirt' and 'it is raining' are both false. But 'I am wearing a red shirt' does not imply 'it is raining', because we can easily think of scenarios where it is true that I am wearing a red shit but it is not raining. — Bram28, Dec 17 '17 at 16:35
@PeterSmith True ... I'll add something to that effect. Thanks! — Bram28, Dec 17 '17 at 16:38
@shuhalo: I've posted an answer, but it does not contradict Bram28's, because Bram28 is pointing out that "$1=1$" is usually provable in pure first-order logic alone, whereas other sentences in mathematics cannot be proven purely logically but only in some formal system that has additional 'non-logical' rules/axioms. So Bram28's remark ties in with what I gave as a second possible interpretation of your question, though in that part of my answer I was focusing on natural numbers for which we believe sentences about them have definite truth value, and yet can never prove all true equivalences. — user21820, Dec 19 '17 at 16:06

score 1 · Answer 3 · answered Dec 19 '17 at 15:59

Actually the answer is yes if your question is interpreted in the most natural way. But it will also make sense once you realize what precisely is meant: $ \def\eq{\Leftrightarrow} $

[Suppose you can prove both $A$ and $B$.]

...

$A$.

$B$.

If $A \land B$ then:

$A \eq B$.

Notice that I did not say anything that is actually different from what you did in your question. I merely emphasized that you first need to be able to prove both $A$ and $B$, and after that can conclude correctly that if $A$ and $B$ are true then $A$ is equivalent to $B$. The non-trivial part of this is precisely in the condition that you must be able to prove $A$ and $B$ before you can prove the equivalence.

Now there is a valid reason you are getting confused. We often think of an equivalence between sentences as something that says that they are equivalent in some 'more fundamental' way. That corresponds to the situation where you can prove $A \eq B$ without proving either of them.

For example, an open problem is whether P=NP. As of today we do not know whether it is true or false, but we can easily prove that (P=NP) is equivalent to ( there exists a polynomial-time program to solve 3SAT. This equivalence is interesting in its own right, because it says that the truth values of those two sentences are necessarily the same.

There is another possible way to interpret your question, in which case the answer would be not quite. Namely if we are talking about sentences $A,B$ over some first-order language that are both true in a structure $M$ that satisfies some theory $T$, it may not be the case that $T$ proves $A \eq B$. After all, $T$ may not 'have enough axioms' to prove it. Worse still, this may be unavoidable for some structures such as the natural numbers with addition and multiplication, because Godel's incompleteness theorem shows (among other stronger facts) that there is no practical first-order theory that proves every true arithmetical sentence! This means that, no matter what foundational system you choose for mathematics, there will be some true arithmetical sentences $A,B$ such that you cannot prove $A \eq B$. It is still true that these true sentences are equivalent (have the same truth value) when applied to the natural numbers, but they are not provably equivalent.

Are all true statements equivalent?

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