# Why not merge the law of excluded middle and the law of noncontradiction with xor?

Consider the following axiom:

A$$_1$$: Either $$P$$ is true, or it isn't; never both

Formally: $$P \oplus \neg P \iff P \lor \neg P \land \neg(P \land \neg P)$$

Contrast this with two of the three axioms of classical logic:

B$$_1$$: Either $$P$$ is true, or it isn't.

B$$_2$$: $$P$$ cannot be true and not true at the same time.

Formally: $$P \lor \neg P, \qquad \neg(P \land \neg P)$$

Now, perhaps the thinking is that A$$_1$$ is actually two axioms in one: it is asserting two things, and thus it isn't an axiom; it's the conjuction of two. However, this thinking is asinine; it is only stating two things because the language requires it to. Using different words, it can easily state one thing:

C$$_1$$: Either $$P$$ is true, xor it isn't.

One statement; one axiom. The difference between one statement and multiple depends on the information packed into the words; the number of statements at the end is just an arbitrary consequence.

Perhaps the thinking is that $$\oplus$$ is a composite logical operator, and that they want the axioms to have formulas that contain only one, non-composite logical operator. Basically, a variation of the above, where the number of logical operators determine the "true" number of statements actually packed within the sentence.

I don't buy this either. Every operator is just a function that relates the individual truths of the atoms to the truth of the entire sentence; this relationship summed up in a truthtable. Xor has one truthtable, just like logical or and logical and.

Certain operators can be used to create the truthtables of other operators, and perhaps the number of such "primitive" operators determine the "true" number of statements. Since or, and and negation are enough to create xor, then the number of xor operators cannot ever give the "true" number of statements. However, why then aren't the axioms formulated from nand, given that this operator, along with negation, can create the truthtables of all the other operators? Is it because nand didn't have a word equivalent in the language the axioms were first made in?

• Heard of truthfuctional completeness... technically (i.e. formally) every connective is just $\land$ and $\lnot$. We just use the shorthands: $\iff, \lor$, etc. for convenience. Commented Jul 16, 2022 at 3:40
• @BertrandWittgenstein'sGhost Yes, I mentioned nand in the Q. Commented Jul 16, 2022 at 4:25
• If you believe in truth tables and view logical operators as functions of truth values, then you're already assuming the axioms you've mentioned, because those axioms are what justify the use of truth tables. The context in which we usually consider these axioms is one where we don't have truth functional semantics, only syntax-based deduction rules. So if you want xor to be "more fundamental" than other connectives, you need to state deduction rules for xor expressions and define the other connectives as shorthand for xor expressions.
– Karl
Commented Jul 16, 2022 at 4:46

When $$A1$$ is the Axiom, $$B1$$ & $$B2$$ are theorems which will be used generally in a variety of theorems, whereas $$A1$$ will be used to (mostly) Prove $$B1$$ & $$B2$$ and then mostly "retire".

When $$B1$$ & $$B2$$ are the Axioms, $$A1$$ will be a theorem which will be used generally unused in other theorems, whereas $$B1$$ & $$B2$$ will continue to be used to Prove other theorems.
Eg 1 Proving P by Contradiction will continue to use $$B2$$ , that is : $$(\neg P \implies P)$$ but $$P$$ & $$\neg P$$ can not be true at the same time.
Eg 2 Proving $$(P \implies Q)$$ & $$(P \equiv Q)$$ will continue to use $$B1$$ , that is : Proving the Individual Case where $$P$$ is true & the Individual Case where $$P$$ is not true.

Here, $$B1$$ & $$B2$$ are more "Powerful" and more general & more useful than $$A1$$, in a sense.

Elaborating a little more with Example :

Axioms must be "Primitive" & "Independent" & "Universal" to be "Powerful".

Consider Natural Numbers with Peano Axioms :
Axiom 1 : 0 is a natural number.
Axiom 8 : For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0.

That is very much better than this odd "combined" Axiom X: 0 is a natural number where (For every natural number n, S(n) = 0 is false). That is, there is a natural number which is not a successor.

Axiom 1 & Axiom 2 are "Primitive" & "Independent" & "Universal" & thus "Powerful".
We might attempt using Axiom X only to Prove Axiom 1 & Axiom 2, but otherwise Axiom X has no use.