Implicational logic confusion

Question

I am having trouble understanding how implicational statements are evaluated to be true or false. Take the statement

• if the 13 axioms of real numbers and $1=0$ are all true, then $1+1 \ne 1$ is true,

  i.e., "the 13 axioms of real numbers and $1=0$" implies "$1+1 \ne 1$".

1. Now, if "the 13 axioms of real numbers and $1=0$" is true, then $1+1 = 1+0 = 1,$ so $1+1 = 1,$ so "$1+1 \ne 1$" is false.

   That is, if "the 13 axioms of real numbers and $1=0$" is true, then "$1+1 \ne 1$" is false.

   This means that the given statement is false.

2. On the other hand, since "the 13 axioms and $1=0$" is false, the given statement, being "false implies "$1+1 \ne 1$", evaluates as true, according to the truth table.

Therefore, the given statement is both true and false. What is my mistake?

Answer 1

• "the 13 axioms of real numbers and $1=0$" implies "$1+1 \ne 1$".

1. Now, if "the 13 axioms of real numbers and $1=0$" is true, then $1+1 = 1+0 = 1,$ so $1+1 = 1,$ so "$1+1 \ne 1$" is false.

   That is, if "the 13 axioms of real numbers and $1=0$" is true, then "$1+1 \ne 1$" is false.

   $\color\red{\text{This means that the given statement is }\textbf{false}.}$

The last line is implicitly assuming that whatever $P$ and $Q$ stand for, $$P\to \lnot Q\tag1$$ and $$P\to Q\tag2$$ have opposite truth values, that is, that they contradict each other. This assumption unfortunately does not hold: as a matter of fact, your otherwise-flawless argument gives a very example of these two sentences being simultaneously true.

Of course, statements $(1)$ and $(2)$ sometimes do have opposite truth values, for example, when $P$ stands for $\text“1=1\text{ or }1\ne1\text”.$ This does not automatically mean that they generally have opposite truth values.

Answer 2

You've shown that a certain thing** --- let's call it $q$ --- implies both $p$ and $!p$. What that tells you is that $q$ is false. There's no mistake in this conclusion, nor any contradiction.

**The "thing" here is "The 13 axioms are true and 1 = 0".

Answer 3

The truth table for implication is given like this:

$P$	$Q$	$P \to Q$
True	True	True
True	False	False
False	True	True
False	False	True

As you can see the full implication is only false when the first part is true and second part is false.

Your first part of the implication "the $13$ axioms of real numbers and $0 = 1$" is false. And as such the whole statement is always true, no matter if the second part is true or false.

How this can be interpreted is that whenever you start with a false claim, any implication from that claim is meaningless. If I say that $0 = 1$, I can easily prove that $1 + 1 = 0$ and that $1 + 1 = 2$. But only one of those is actually true given normal rules of mathematics.