what is the intuitive explanation of the truth of the statement if 0=1, then 1=2?

Question

Consider the statement if 0=1, then 1=2 This statement is weird but since the hypothesis (if 0=1) is false, then according to propositional logic the entire statement is true, although the conclusion (then 1=2) is false. I just cannot understand how can we consider the entire statement to be true even though our conclusion is wrong. Can you provide another examples where such situation is easy to grasp? Maybe I can get it with concrete examples.

Answer 1

We never satisfy the hypothisis thus we can write whatever we want as a conclusion. Such a statment is called a "vacuous truth."

As for more concrete examples: If the pigs fly, then blue is brown. If Toronto is in Germany, then 1=2.

Answer 2

(This is really expanding on Noah Schweber's comment.)

We have a true rule of arithmetic, that if $a = b$ then $a+1 = b+1$. (This follows in any system where equal things may be substituted for each other.)

So if $0 = 1$, then $0+1 = 1+1$, so $1 = 2$. You applied a perfectly valid transformation ("add $1$ to both sides") to an input, validly obtaining an output. The fact that the input was nonsense means the output was nonsense, but that doesn't make the transformation bad.

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An implication $A \to B$ can be thought of as a process that takes an $A$ and returns a $B$. The implication "$0 = 1 \to 1 = 2$" is a process that takes a proof that $0 = 1$, transforms it (e.g. by appending the line "you can add $1$ to both sides of an equality"), and outputs a proof that $1 = 2$. The process is valid, even though you can never actually execute it validly because you will never be able to provide it with a valid input (i.e. a proof that $0 = 1$).

Answer 3

Here's an analogy that might be useful if you have any coding experience. If you have a conditional statement "if $P$ then $Q$," the compiler checks to see if $P$ is true before proceeding. If it is not, the compiler moves on without executing $Q.$ So the statement executes without an error.

Answer 4

I cannot understand how can we consider the entire ‘if-then’ statement to be true even though its conclusion is false.

1. Notice that a conditional $$A\to B\tag1$$ and its consequent $$B\tag2$$ have different truth values only when $A$ and $B$ are both false. So, essentially, you are asking why sentences $(1)$ and $(2)$ aren't just equivalent to each other. But why should they be? After all, for example, the truth of $$A \land B$$ is determined by considering the truths of both $A$ and $B$ rather than the truth of just $B.$ Now, since sentence $(1)$ is logically equivalent to $$\lnot A\lor B$$ (they have the same truth table), it certainly can be true when $B$ is false.

2. A more satisfying and illuminating explanation of your discomfort is that because we commonly frame conditional statements as

   • if premise/condition, then conclusion/implication

   referring to an “implication” variously as the entire conditional $(1)$ as well as just its consequent $(2)$ (!), we tend to carelessly forget that they aren't actually equivalent to each other.

Answer 5

The word "if" means something different in mathematical English than it means in ordinary English. In mathematical English, the word "if" is an abbreviation for a particular statement involving the words "not" and "or."

Specifically, when a mathematician writes a sentence like

if p, then q,

then what they are actually saying is that

either it's false that p, or it's true that q.

So, when a mathematician writes the sentence "if $0 = 1$, then $1 = 2$," then what the mathematician is actually claiming is:

either it's false that $0 = 1$, or it's true that $1 = 2$.

This claim is, of course, true.

Answer 6

We have the tautology: $ A \implies [\neg A \implies B]~$ (the principle of vacuous truth)

Source: https://www.erpelstolz.at/gateway/TruthTable.html

In your case, $A ~\equiv~ 0\neq 1$ and $B ~\equiv~ 1=2$:

$~~~~~~0\neq 1 \implies [0=1 \implies 1=2]$

When $A$ is true (lines 1 and 2 of the truth table), the implication of $\neg A \implies B$ will be true regardless of the truth value of $B$. In that case, we cannot use this result to infer anything about the truth value of $B$.

Using a form of natural deduction we can prove this tautology as follows (screenshot from my proof checker):

Answer 7

There are several answers here that address the technical content of your question. I want to add some words about the psychological content.

When you use an implication like $A \implies B$ in ordinary English or mathematical discourse you usually intend to express some kind of causality. Psychologically, you are saying that the truth of $A$ somehow causes or forces $B$ to be true. That's what the argument "proving" the implication demonstrates. When $A$ is false it tells you nothing about $B$.

Your problem arises when you try to define the meaning of the implication "$A \implies B$" formally. The usual way to do that is with truth tables, which leads to the true implication

The moon is made of green cheese $\implies$ I am the King of England.

I'm not the King of England, but the implication is correct.

In general terms, if $A$ is false then $A \implies B$ is true.

That definition of implication confuses many students. It's introduced to make everyday logic more formal and mathematical, but it's very rarely relevant when beginners construct real mathematical proofs. I think it's not a good way to introduce logic and proof.

Logicians and philosophers struggle with it too, at a formal level. See https://www.britannica.com/topic/implication .