I hold that mathematics is based on tautologies. That is why you can come up with mathematical equations that do not refer to anything in the physical universe. It’s kind of like the CTEK theory of justification in that it just circles back on itself. Now, I don’t know anything about Peano arithmetic, but the Arabic numerals normal people use as in this example is a tautology.

A Tautology is a feature of a language, not a theory. Peano arithmetic is a theory, so something being true in Peano arithmetic does not make it a tautology of the language in which it is written. In the very same language the following are correct:

1+1=0

in binary arithmetic, and

1+1=1

in Boolean arithmetic.

Boolean arithmetic and binary arithmetic ARE languages as are all mathematics.

No they are theories, expressed in the language of mathematics. The distinction is critical.

Please explain the critical distinction. A theory suggests that it’s foundation is in the observable physical universe. Or do you disagree?

Look up any logic text. There is no shame in not knowing the difference between a theory and a language. Most people don't. But it is inappropriate to criticise other posts in a discussion on logic, based solely on your lack of knowledge of some of the basic building blocks of symbolic logic.

Just humor me a bit. Does the theory refer to anything in physical reality, or is it free-floating? This is critical to my understanding. If it is free-floating, then I stand by my claim.

“1+1=2” is a mathematical tautology that is self-evident to anyone who ever had an apple then was given another one. Theories of logic just seem like convoluted language gaming to me. Perhaps you could explain to someone ignorant (as I am) why or how a theory can’t be based on tautologies.

Hello. Can you please state what you understand to be a tautology?

If someone asks a mathematician or logician to prove "1+1=2", they will do so using axioms and theorems of arithmetic. They will not take it as self-evident.

I understand a tautology to be a proposition that has to be true given the meanings of the terms used. Is that wrong?

Hi. You could try breaking the implication into smaller expressions and treating them as an argument. You have 3 main conjuncts in the antecedent; treat them as your premises.
Treat your consequent as the conclusion you want to prove using the 3 premises.

First let me write it out in a way that's a bit easier to read - for me at least:

"If exactly one object has property F and exactly one object has property G and no object has both properties, then there are exactly two objects that satisfy one or more of the two properties"

It sounds like it should be true.

I skipped symbolic logic in college because I didn’t need it to graduate. The joke about nerds and getting laid was just that, a joke.

The idea of mathematical tautologies was suggested to me by my professor who got his PhD from UCLA.

Very roughly, a formal language is (1) a set of symbols that can be used, together with (2) a set of rules for how they can be strung together to make syntactically correct statements and (3) a set of rules for how deductions can be made.

A theory T in a language L is a set of syntactically valid statements in L such that any statement that can be deduced from a finite collection of statements in T, using the deduction rules of L, is also in T.

We say a collection A of statements in L is a set of 'axioms' for T if T is the intersection of all theories containing A. We say 'T is generated by A'.

The set of tautologies in L is the theory generated by the empty set (ie no axioms).

I re expressed some of the formulas. Proving the conclusion using the 3 premises would be equivalent to proving that the initial implication is true.



P1: ∃x(Fx ∧ ∀y(Fy ⊃ y=x) )
P2: ∃x(Gx ∧ ∀y(Gy ⊃ y=x) )
P3: ∀x(¬Fx ∨ ¬Gx )

C1: ∃x∃y[ (
( (Fx ∨ Gx) ∧ (Fy ∨ Gy) )
∧ (x≠y) )
∧ ∀z((Fz ∨ Gz) ⊃ (z=x ∨ z=y) )]

The set of tautologies in L is the theory generated by the empty set (ie no axioms). — andrewk

I don’t understand what you mean by this.

How do you define “tautology”?

Rather than wasting more of your time, perhaps you could suggest to me a symbolic logic textbook that I can download to my Kindle? Thanks.

It's a good question, often asked and unfortunately one for which I do not have an answer, as I did not use textbooks in learning logic. It would be great if we had a list of recommended textbooks pinned to the top of the Logic and Mathematics forum, so that all could benefit from it. I'll start a thread there inviting recommendations. If we get some good ones (ie recommendations that don't have lots of other people saying the text is terrible), we can pin it.

That’s a good idea. Perhaps if one is popular, then we could do a reading group?

Yes that sounds good. I have started a thread asking for recommendations.

cool

Assume that F and G are "attributes". Assume that beings have some combination of attributes.
the number of beings with an attribute combination which contains F or G is sum over all possible attribute combinations which contain F or G, of the number of beings with that combination.

The only possible attribute combinations which contain F or G are of the types
1) F not G
2) G not F
3) F and G

Define Nf as the number of beings with F only. Define Ng as the number of beings with G only. Define Nf&g as the number of beings with F and G. Define Nf|g as the number of beings with at least one of F or G. Then

Nf|g = Nf + Ng + Nf&g

for the special case that Nf = 1, Ng = 1, and Nf&g = 0, Nf|g=2

Here you go:

P1: ∃x(Fx ∧ ∀y(Fy → y=x) )
P2: ∃x(Gx ∧ ∀y(Gy → y=x) )
P3: ∀x¬(Fx ∧ Gx )

C1: ∃x∃y[ (
(Fx ∨ Gx)
∧ (Fy ∨ Gy)
∧ (x≠y) )
∧ ∀z((Fz ∨ Gz) → (z=x ∨ z=y) )]

We write ‘a’ for an object that satisfies P1 and ‘b’ for an object that satisfies P2.
So we have:

3: Fa ∧ ∀y(Fy → y=a) (∃ elimination, P1)
4: Gb ∧ ∀y(Gy → y=b) (∃ elimination, P2)

which we split into

5: Fa
6: ∀y(Fy → y=a)
7: Gb
8: ∀y(Gy → y=b)

9: ¬(Fa ∧ Ga ) (substitution of a into P3)
10: ¬(Fb ∧ Gb ) (substitution of b into P3)

Next:
11: ....Fb (Cond Hyp)
12: ....Fb ∧ Gb (9, 7)
13: ....Contradiction (10, 12)
14: ¬Fb (close Cond Proof, negating 11)

15: ....Ga (Cond Hyp)
16: ....Fa ∧ Ga (15, 5)
17: ....Contradiction (9, 16)
18: ¬Ga (close Cond Proof, negating 15)

We hypothesise that a and b are objects that satisfy the existence claim in C1.
We’ll prove the conjuncts in C1 in turn, for the case x=a, y=b.

19: Fa ∨ Ga (5, OR introduction) [1st conjunct proven]
20: Fb ∨ Gb (7, OR introduction) [2nd conjunct proven]

21: ....a=b (Cond Hyp)
22: ....Ga (7, 21, substitution on =)
23: ....Fa ∧ Ga
24: ....¬(Fa ∧ Ga) (P3, specification of x=a)
25: ....Contradiction (23, 24)
26: a ≠ b (close Cond Proof, negating 21) [3rd conjunct proven]


27: Fz → z=a (specify y=z in 6)
28: z ≠ a → ¬Fz (reversal of 27)
29: Gz → z=b (specify y=z in 8)

31: .... Fz ∨ Gz (Cond Hyp)
32: ........ z ≠ a (Cond Hyp)
33: ........ ¬Fz (Modus Ponens 28, 32)
34: ........ Gz (31, 33, OR elimination)
35: ........ z=b (MP 29, 34)
36: .... ¬ (z=a) → (z=b) (close Cond Proof 32-35)
37: .... z=a ∨ z=b
38: Fz ∨ Gz → z=a ∨ z=b (close Cond Proof 31-37) [4th conjunct proven]

39: (Fa ∨ Ga) ∧ (Fb ∨ Gb) ∧ a ≠ b ∧ (Fz ∨ Gz → z=a ∨ z=b) (AND introduction 19, 20, 26, 38)
40: ∃x ∃y (Fx ∨ Gx) ∧ (Fy ∨ Gy) ∧ x ≠ y ∧ (Fz ∨ Gz → z=x ∨ z=y) (∃ introduction 39)

Slightly different proof:


1- Fx' ∧ ∀y(Fy ⊃ y=x') Existential instantiation (P1)
2- Gy' ∧ ∀y(Gy ⊃ y=y') Existential instantiation (P2)
3- Fx' Simplification (1)
4- ∀y(Fy ⊃ y=x') Simplification (1)
5- Gy' Simplification (2)
6- ∀y(Gy ⊃ y=y') Simplification (2)
7- ¬Fx' ∨ ¬Gx' Universal instantiation (P3)
8- ¬Fy' ∨ ¬Gy' Universal instantiation (P3)
9- ¬Gx' Disjunctive syllogism (3,7)
10- ¬Fy' Disjunctive syllogism (5,8)
11- Fx' ∧ ¬Fy' Adjunction (3,10)
12- ¬(x' = y') Identity (11)
13- Fx' ∨ Gx' Addition (3)
14- Fy' ∨ Gy' Addition (5)
15- ∀y(Fy ⊃ y=x' ∨ y=y') Addition (4)
16- ∀y(Gy ⊃ y=x' ∨ y=y') Addition (6)
17- ∀y(Fy ⊃ y=x'∨y=y') ∧ ∀y(Gy ⊃ y=x' ∨ y=y') Adjunction (15,16)
18- ∀y[(Fy ⊃ y=x'∨y=y') ∧ (Gy ⊃ y=x'∨y=y')] Universal distribution (17)
19- ∀y[(¬Fy ∨(y=x'∨y=y')) ∧ (¬Gy ∨ (y=x'∨y=y'))] Implication (18)
20- ∀y[(¬Fy ∧ ¬Gy) ∨ (y=x'∨y=y')] Distribution law (19)
21- ∀y[(Fy ∨ Gy) ⊃ (y=x'∨y=y')] Implication (20)
C - ∃x∃y[ ( ( (Fx ∨ Gx) ∧ (Fy ∨ Gy) ) ∧ (x≠y) )
∧ ∀z((Fz ∨ Gz) ⊃ (z=x ∨ z=y) )] Existential Generalization (12,13,14,21)

I didn't read any of this thread yet but if I had to prove that 1 + 1 = 2 I'd start from the Peano axioms and define $n + 1 \overset{\text{def}}{\equiv} S(n)$where $S$ is the successor function given by the axioms. The symbol 0 is also given by the axioms. Then I'd define the symbols:

$1 \overset{\text{def}}{\equiv} S(0) = 0 + 1$

and

$2 \overset{\text{def}}{\equiv} S(1) = 1 + 1$

and I'm done.

It turns out the thread is actually about a completely different problem. The title is a misnomer.

That's what I'm talking about! I just woke up, so I'll do some things and se your reply calmy later. Thanks for answering!
Yeah, as I said before, I just put this title because the book says that sentence is the representation of 1+1=2, but it isn't really important to what I was proposing. The title was just something that i thought it would call people's attention; maybe it's inadequate.

I take it, from the way this post has developed, none of the contributors is actually familiar with the Peano Axioms of arithmetic. Firstly, the fact that 1+1=2 can be proved, means that it is a theorem, and therefore MUST be proved - ie, it is not an axiom. Secondly, the proof that 1+1=2 is actually much simpler than they imagine. In its short form, it goes as follows:

1+1 = 1+S(0) = S(1+0) = S(1) = 2

- where the S function = "the successor of".

In its long form, of course, it would require an exposition of the Peano Axioms, which goes beyond the scope of this post.

I take it, from the way this post has developed, none of the contributors is actually familiar with the Peano Axioms of arithmetic. — alan1000

You mean aside from the answers by and ?