# Axiomatics 2

#### solakis

##### Active member
Given the following system of axioms

For all A,B,C:

1) A+B=B+A

2) A+(B+C) =(A+B)+C

3) A.B=B.A

4) A.(B.C) = (A.B).C

5) A.(B+C)= A.B+A.C

6) A+0=A

7) A.1=A

8) A+(-A)=1

9) A.(-A)=0

10) A+(BC) = (A+B).(A+C)

11) $$\displaystyle 1\neq 0$$

Then prove :

If A+B=0 then A=0 AND B=0

#### Barioth

##### Member
Re: axiomatics 2

I don't have time to solve it all, but maybe (Just maybe...) this can help.

$$\displaystyle A+1=A+1*1=(A+1)(A+1)=AA+A+A+1$$ we've used A7,A5 and A10
$$\displaystyle A+1=AA+A+(A+1) => AA+A=0$$ A6

so now we have AA+A= 0, we'll call this C1

$$\displaystyle 0=A+B=A+B*1=(A+B)(A+1)=AA+BA+A+B=BA+B$$ We've used Hyp, A7,A10, A5 and C1.

$$\displaystyle BA+B=0 => B=A$$ from C1

Now you must show that if $$\displaystyle A+A=0$$ then $$\displaystyle A=0$$

Note: I'm a bit tirred, I may have made a mistake.. (I hope not, but you know how it is..)

edit: Those axioms aren't the one you use for real, I take for granted, this exercise is for fun only.

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#### solakis

##### Active member
Re: axiomatics 2

I don't have time to solve it all, but maybe (Just maybe...) this can help.

$$\displaystyle A+1=A+1*1=(A+1)(A+1)=AA+A+A+1$$ we've used A7,A5 and A10
$$\displaystyle A+1=AA+A+(A+1) => AA+A=0$$ A6

so now we have AA+A= 0, we'll call this C1

$$\displaystyle 0=A+B=A+B*1=(A+B)(A+1)=AA+BA+A+B=BA+B$$ We've used Hyp, A7,A10, A5 and C1.

$$\displaystyle BA+B=0 => B=A$$ from C1

Now you must show that if $$\displaystyle A+A=0$$ then $$\displaystyle A=0$$

Note: I'm a bit tirred, I may have made a mistake.. (I hope not, but you know how it is..)

edit: Those axioms aren't the one you use for real, I take for granted, this exercise is for fun only.

You have proved that :

A+1=0,but A+1 = 1 so you proved 1=0
But by axiom 11 we have :$$\displaystyle 1\neq 0$$

AA+BA+A+B=AA+BA

A.A=A,

A+A=A

Nothing is for fun in mathematics.

The above system is a boolean system

Thanks for the help anyway.

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#### Barioth

##### Member
Re: axiomatics 2

Nothing is for fun in mathematics.
Then I feel sorry for you...

The above system is a boolean system
This help a lot, next time, say so...

we have that A is 0 or 1 since A is a boolean.

we have 0+0 =0 from A6

we have 1+0=0+1 = 1 from A6 and A1

we only need to evalute 1+1 now

but we can get
A+1= (A+1)*1 from A7
(A+1)*1= (A+1)*(A+(-A)) from A9
(A+1)*(A+(-A))= A+(-A*1)=A+(-A) from A10 and A7
A+(-A)=1 from A8
so we have A+1 = 1 so 1+1=1.

#### solakis

##### Active member
Re: axiomatics 2

Then I feel sorry for you...

we have that A is 0 or 1 since A is a boolean.

.
This is not always true:

For example if A,B,C are subsets of a set V ,with the well known operations of union intersection and complimentation,and with constants V and the empty set ,then we have a boolean system.

And definetelly we cannot say that each set is empty or V.

Hence your proof is not general.

It may have an application on the boolean system of propositional calculus

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Given the following system of axioms

For all A,B,C:

1) A+B=B+A
2) A+(B+C) =(A+B)+C
3) A.B=B.A
4) A.(B.C) = (A.B).C
5) A.(B+C)= A.B+A.C
6) A+0=A
7) A.1=A
8) A+(-A)=1
9) A.(-A)=0
10) A+(BC) = (A+B).(A+C)
11) $$\displaystyle 1\neq 0$$

Then prove :
If A+B=0 then A=0 AND B=0
Lemma L1: A.A = A
Proof: A = A.1 = A.(A+(-A)) = AA+A(-A) = AA+0 = AA. [using A7, A8, A5, A9, A6]

Lemma L2: A.0 = 0
Proof: A.0 = A.(A.(-A)) = (A.A).(-A) = A.(-A) = 0. [using A9, A4, L1, A9]

Suppose A≠0, B=0, and A+B=0.
Then A+B = A+0 = A ≠ 0. [using A6]

Suppose A=0, B≠0, and A+B=0.
Then A+B = 0+B = B+0 = B ≠ 0. [using A1, A6]

Suppose A≠0, B≠0, and A+B=0 (H1).
Then A = A+0 = A+B.0 = (A+B).(A+0) = 0.(A+0) = (A+0).0 = 0. [using A6, L2, A10, H1, A3, L2]

Therefore if A+B=0, then A=0 AND B=0.

#### solakis

##### Active member
Lemma L1: A.A = A
Proof: A = A.1 = A.(A+(-A)) = AA+A(-A) = AA+0 = AA. [using A7, A8, A5, A9, A6]

Lemma L2: A.0 = 0
Proof: A.0 = A.(A.(-A)) = (A.A).(-A) = A.(-A) = 0. [using A9, A4, L1, A9]

Suppose A≠0, B=0, and A+B=0.
Then A+B = A+0 = A ≠ 0. [using A6]

Suppose A=0, B≠0, and A+B=0.
Then A+B = 0+B = B+0 = B ≠ 0. [using A1, A6]

Suppose A≠0, B≠0, and A+B=0 (H1).
Then A = A+0 = A+B.0 = (A+B).(A+0) = 0.(A+0) = (A+0).0 = 0. [using A6, L2, A10, H1, A3, L2]

Therefore if A+B=0, then A=0 AND B=0.
I am not so sure .

1) What law of logic make you to consider those three cases

2) Again what law of logic justifies the conclusion corresponding to those three hypothesis

3) What allows you to conclude : $$\displaystyle A+B\neq 0$$ if A+B=A and $$\displaystyle A\neq 0$$

#### Barioth

##### Member
mhm, this problem is pretty interesting, but shoudn't make so much touble...

#### Deveno

##### Well-known member
MHB Math Scholar
I am not so sure .

1) What law of logic make you to consider those three cases
The "law of the excluded middle":

In any case we are given that A+B = 0.

Either B = 0, or not. If B = 0, then either A = 0, or not.

If B = 0 and A = 0, there is nothing to prove, we already have the desired conclusion. Otherwise, we have B = 0, and A ≠ 0, which is case 1.

On the other hand if B ≠ 0, then either A = 0, or not. The first case is case 2, the latter is case 3.

2) Again what law of logic justifies the conclusion corresponding to those three hypothesis
The axioms you listed in your first post, which are taken to hold for ALL elements of the system under consideration.

3) What allows you to conclude : $$\displaystyle A+B\neq 0$$ if A+B=A and $$\displaystyle A\neq 0$$
Let's go step-by-step through case (1):

Assumptions: A ≠ 0, B = 0, A + B = 0.

Since B = 0 (that is B is identical to 0), the expression A + B is the same as the expression A + 0. This is called "substituting the value 0 in for B", which we can do because they are EQUAL.

Now A + 0 = A, by axiom 6.

So A + B = A + 0 = A, thus A + B = A, by transitivity of equality.

Since A ≠ 0, and A + B = A, we must have A + B ≠ 0, for if we had

A + B = 0, then:

0 = A + B (by symmetry of equality)

A + B = A, and thus:

0 = A (by transitivity of equality), and thus A = 0 (by symmetry of equality).

But this violates our original assumption that A ≠ 0:

something cannot be both equal and not equal to something else.

This is another consequence of the law of the excluded middle, which can be stated as either:

P or not-P = TRUE (tautological form)

or:

P and not-P = FALSE (contradiction form).

******

In mathematics, a bivalent system of truth based on a boolean system of logic (satisfying the above axioms you gave with:

1 <--> "true"
0 <--> "false"
+ <--> "or"
. <--> "and"
- <--> "not")

is usually taken as a GIVEN, unless you are a constructivist (or intuitionist), in which case a DIRECT proof must be exibited.

Let's see if a direct proof is possible for this theorem (using the axioms given).

By A1, it is sufficient to prove that:

A + B = 0 implies A = 0, since we can just switch the roles of A and B to prove that:

A + B = 0 implies B + A = 0 implies B = 0, by the proof we are about to demonstrate:

A + B = 0 (given)

(A + B).1 = 0 (from A7)

(A + B).(A + (-A)) = 0 (from A8)

A + (B.(-A)) = 0 (from A10)

A + ((B.(-A)) + 0) = 0 (from A6, applied to B.(-A))

A + ((B.(-A)) + (A.(-A))) = 0 (from A9)

A + (((-A).B) + ((-A).A)) = 0 (using A3 twice, on B.(-A) and A.(-A))

A + ((-A).(B + A)) = 0 (using A5, applied to ((-A).B) + ((-A).A))

A + ((-A).(A + B)) = 0 (using A1, applied to B + A)

A + ((-A).0) = 0 (using our given A + B = 0)

A + 0 = 0 (using L2 proved by ILikeSerena, applied to (-A).0)

A = 0 (using A6).

#### solakis

##### Active member
The "law of the excluded middle":

In any case we are given that A+B = 0.

Either B = 0, or not. If B = 0, then either A = 0, or not.

If B = 0 and A = 0, there is nothing to prove, we already have the desired conclusion. Otherwise, we have B = 0, and A ≠ 0, which is case 1.

On the other hand if B ≠ 0, then either A = 0, or not. The first case is case 2, the latter is case 3.

The axioms you listed in your first post, which are taken to hold for ALL elements of the system under consideration.

Let's go step-by-step through case (1):

Assumptions: A ≠ 0, B = 0, A + B = 0.

Since B = 0 (that is B is identical to 0), the expression A + B is the same as the expression A + 0. This is called "substituting the value 0 in for B", which we can do because they are EQUAL.

Now A + 0 = A, by axiom 6.

So A + B = A + 0 = A, thus A + B = A, by transitivity of equality.

Since A ≠ 0, and A + B = A, we must have A + B ≠ 0, for if we had

A + B = 0, then:

0 = A + B (by symmetry of equality)

A + B = A, and thus:

0 = A (by transitivity of equality), and thus A = 0 (by symmetry of equality).

But this violates our original assumption that A ≠ 0:

something cannot be both equal and not equal to something else.

This is another consequence of the law of the excluded middle, which can be stated as either:

P or not-P = TRUE (tautological form)

or:

P and not-P = FALSE (contradiction form).

******

In mathematics, a bivalent system of truth based on a boolean system of logic (satisfying the above axioms you gave with:

1 <--> "true"
0 <--> "false"
+ <--> "or"
. <--> "and"
- <--> "not")

is usually taken as a GIVEN, unless you are a constructivist (or intuitionist), in which case a DIRECT proof must be exibited.

Let's see if a direct proof is possible for this theorem (using the axioms given).

By A1, it is sufficient to prove that:

A + B = 0 implies A = 0, since we can just switch the roles of A and B to prove that:

A + B = 0 implies B + A = 0 implies B = 0, by the proof we are about to demonstrate:

A + B = 0 (given)

(A + B).1 = 0 (from A7)

(A + B).(A + (-A)) = 0 (from A8)

A + (B.(-A)) = 0 (from A10)

A + ((B.(-A)) + 0) = 0 (from A6, applied to B.(-A))

A + ((B.(-A)) + (A.(-A))) = 0 (from A9)

A + (((-A).B) + ((-A).A)) = 0 (using A3 twice, on B.(-A) and A.(-A))

A + ((-A).(B + A)) = 0 (using A5, applied to ((-A).B) + ((-A).A))

A + ((-A).(A + B)) = 0 (using A1, applied to B + A)

A + ((-A).0) = 0 (using our given A + B = 0)

A + 0 = 0 (using L2 proved by ILikeSerena, applied to (-A).0)

A = 0 (using A6).
O.K lets take your proof step by step:

How do you go from : A+B=0 to (A+B).1=0??

Because A7 tells you that : A.1 =A

And certainly does not tell you how to go from A+B=0 to (A+B).1 =0

#### Deveno

##### Well-known member
MHB Math Scholar
A + B = 0 (given)

(A + B).1 = A + B (A7, which applies to ANY element of our system, including A+B)

(A + B).1 = 0 (transitivity of equality).

The axiom:

A.1 = A for all A

doesn't mean A.1 = A "just for A", but rather:

A.1 = A
B.1 = B
C.1 = C
X.1 = X, where X is any legal expression formed from elements {A,B,C,...etc.} and the symbols +,. and - (which are binary (the first two) and unary (the last one) operations), perhaps appropriately grouped with parentheses.

Your axiom system is a bit abbreviated, it should mention an underlying set of elements, the fact that + and . are binary operations, that - is a unary operation, that there are two distinguished elements of our underlying set 0 and 1 with special properties (defined by axioms 6,7,8,9 and 11). However these things are "standard" when studying axiomatic systems.

So formally, we have a system:

U = {S,+,.,-,1,0}

where + is a function

SxS --> S

and . is a function

SxS -->S

and - is a function

S-->S

and 1,0 are elements of S

such that axioms 1 through 11 hold for all elements of S.

The fact that - is a function S-->S is often stated instead like so:

For every A in S, there is a unique B in S such that:

A + B = 1 and A.B = 0

in which case B is denoted -A.

The uniqueness requirement is often omitted, however, because if:

A + B = A + C = 1 and
A.B = A.C = 0

Then (-A).(A + B) = (-A).(A + C) so:

(-A.A) + (-A.B) = (-A.A) + (-A.C)

0 + (-A.B) = 0 + (-A.C)

-A.B = -A.C

and from

A.B = A.C = 0, we have:

A.B + (-A.B) = A.C + (-A.C)

(A + (-A)).B = (A + (-A)).C

1.B = 1.C

B.1 = C.1

B = C

I leave it as an exercise for you to consider which axioms justify each step above.

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#### solakis

##### Active member
A + B = 0 (given)

(A + B).1 = A + B (A7, which applies to ANY element of our system, including A+B)

(A + B).1 = 0 (transitivity of equality).
But the rule of transitivity says:

IF X=Y and Y=Z Then X=Z

So if we put : X= (A+B).1 , Y= A+B, Z=0

WE have :

IF (A+B).1 =A+B and A+B= 0 ,THEN (A+B).1=0

And now how do we get: (A+B).1 =0??

#### Deveno

##### Well-known member
MHB Math Scholar
(A+B).1 = A+B is guaranteed by axiom 7.

A+B = 0 is given as a premise of the problem.

So the "if" part is satisfied, and thus we can conclude the "then" part.

#### solakis

##### Active member
(A+B).1 = A+B is guaranteed by axiom 7.

A+B = 0 is given as a premise of the problem.

So the "if" part is satisfied, and thus we can conclude the "then" part.

for example if we say :

IF 2+2=4 THEN 3+2=6

The "if" part is satisfied because 2+2=4
Can we conclude 3+2=6 ??

#### Deveno

##### Well-known member
MHB Math Scholar
The "if...then" statement known as the transitivity of equality is a true statement. It's one of several statements in mathematics that are taken as "axiomatically true" or "true by definition".

If 2 +2 = 4 then 3 + 2 = 6

is logically false, because while 2+2 = 4 is indeed true, 3 + 2 = 6 is not:

T-->T = T
T-->F = F

(statements of the form F-->anything are typically not considered in math, because anything can be proven from a false premise, so no useful knowledge is revealed).

The transitivity of equality:

If A = B and B = C, then A = C

is taken to be true A PRIORI. In ordinary language, this translates to the idea:

If thing 1 is exactly the same thing as thing 2, and thing 2 is exactly the same thing as thing 3, then all three things are the same thing.

In practice, we use such transitivity to establish such things as:

10 = 5 + 5
5 + 5 = 4 + 6

therefore, 10 = 4 + 6.

If you are uncomfortable with the reflexive, symmetric and transitive nature of equality, I suggest you give up on the pursuit of mathematics entirely, as you will no doubt encounter many grave difficulties.

Just to be clear, I am not suggesting ANY if...then clause can be verified solely by verifying the "if" part, but rather, any INSTANCE of transitivity can be justified by showing the antecedent clause of the rule of transitivity is satisfied, thus allowing us to safely (by invoking the rule of transitivity) deduce the consequent.

A bit more formally (but perhaps more mystifying):

1) $\forall X,Y,Z: [(X = Y)\wedge(Y = Z)] \implies (X = Z)$ (given, rule of transitivity)
2) $X = Y$ (given)
3) $Y = Z$ (given)
4) $(X = Y)\wedge(Y = Z)$ (conjunction introduction, 2,3)
5) $X = Z$ (modus ponens, 1,4)

By contrast, we cannot derive B from A, given only that A is true (your argument): both A AND A-->B must BOTH be true.

#### solakis

##### Active member
The "if...then" statement known as the transitivity of equality is a true statement. It's one of several statements in mathematics that are taken as "axiomatically true" or "true by definition".

If 2 +2 = 4 then 3 + 2 = 6

is logically false, because while 2+2 = 4 is indeed true, 3 + 2 = 6 is not:

T-->T = T
T-->F = F

(statements of the form F-->anything are typically not considered in math, because anything can be proven from a false premise, so no useful knowledge is revealed).

The transitivity of equality:

If A = B and B = C, then A = C

is taken to be true A PRIORI. In ordinary language, this translates to the idea:

If thing 1 is exactly the same thing as thing 2, and thing 2 is exactly the same thing as thing 3, then all three things are the same thing.

In practice, we use such transitivity to establish such things as:

10 = 5 + 5
5 + 5 = 4 + 6

therefore, 10 = 4 + 6.

If you are uncomfortable with the reflexive, symmetric and transitive nature of equality, I suggest you give up on the pursuit of mathematics entirely, as you will no doubt encounter many grave difficulties.

Just to be clear, I am not suggesting ANY if...then clause can be verified solely by verifying the "if" part, but rather, any INSTANCE of transitivity can be justified by showing the antecedent clause of the rule of transitivity is satisfied, thus allowing us to safely (by invoking the rule of transitivity) deduce the consequent.

A bit more formally (but perhaps more mystifying):

1) $\forall X,Y,Z: [(X = Y)\wedge(Y = Z)] \implies (X = Z)$ (given, rule of transitivity)
2) $X = Y$ (given)
3) $Y = Z$ (given)
4) $(X = Y)\wedge(Y = Z)$ (conjunction introduction, 2,3)
5) $X = Z$ (modus ponens, 1,4)

By contrast, we cannot derive B from A, given only that A is true (your argument): both A AND A-->B must BOTH be true.
So:
A)The argument:
If (A+B).1=A+B and A+B=0,then (A+B).1=0,and since (A+B).1=A+B and A+B=0
We have :
(A+B).1=0
Is correct
Whilst
The argument:
If 2+2=4,then 2+3=5 and since 2+2=4
We have:
2+3=5
Is not correct

B) According to the last part of your post we have:
1)A+B=0…………………………………………………………given
2)(A+B).1 = A+B…………………………………………………axiom A.1=A
3)(A+B).1 = A+B and A+B=0 => (A+B).1=0…………………………transitivity property
4) )(A+B).1 = A+B and A+B=0…………………………………………..1,2 Addition property
5) (A+B).1=0………………………………………………………………………3,4 M.ponens
Can you do the rest of the proof in that style ??

#### Deveno

##### Well-known member
MHB Math Scholar
I probably could, but if you really want a formal style proof Evgeny.Makarov is a much better person for that sort of thing. I lean more towards algebraic arguments, not logical propositions.

#### solakis

##### Active member
I probably could, but if you really want a formal style proof Evgeny.Makarov is a much better person for that sort of thing. I lean more towards algebraic arguments, not logical propositions.
What does a algebraic argument consists of

What is a logical proposition

Pardon me but i am so sure yet while my argument :

If 2+2=4,then 2+3=6,but since 2+2=4 ,then 3+3=6

is correct or not

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar

for example if we say :

IF 2+2=4 THEN 3+2=6

The "if" part is satisfied because 2+2=4
Can we conclude 3+2=6 ??
Yes, we can. This argument is valid, but it is not sound because the premise "IF 2+2=4 THEN 3+2=6" is not true.

#### solakis

##### Active member
Yes, we can. This argument is valid, but it is not sound because the premise "IF 2+2=4 THEN 3+2=6" is not true.
Are the following arguments valid and sound:

if 2+3=6 then 2+2=4,therefor 2+2=4

If i go to London i will buy a car,therefor i will buy a car

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Are the following arguments valid and sound:

if 2+3=6 then 2+2=4,therefor 2+2=4
There are two options regarding validity. If we insist on deriving 2+2=4 from "If 2+3=6 then 2+2=4", i.e., if the argument is complete and there is no way to derive 2+2=4 using some other reasoning that is not shown, then the argument is not valid. It has the form "$A\to B$, $B$", and this transition does not conform to any of the inference rules. Modus Ponens could be used to derive $B$ from $A\to B$, but then the argument must also contain a proof of $A$.

The second option is that the argument is abbreviated and 2+2=4 is derived not from "If 2+3=6 then 2+2=4", but using other means that are not shown; for example, 2+2=4 is an axiom. Then the argument is valid. The premise "If 2+3=6 then 2+2=4" is simply not used.

If i go to London i will buy a car,therefor i will buy a car
This argument is either incomplete or invalid. Again, there are two options: either you derive "I will buy a car" essentially using the premise "If I go to London, I will buy a car", or you derive it without using that premise, using some independent means. In the first case, you also need to prove "I will go to London" in order to use Modus Ponens. In the second case, the argument is incomplete, and without extra information there is less reason to think that you will buy a car than to believe that 2 + 2 = 4.

#### Deveno

##### Well-known member
MHB Math Scholar
I'd like to point out that statements like:

"IF 2+3 = 6, THEN 2+2 = 4"

are usually not considered within mathematical study.

This is actually a TRUE statement, because it can be proved by other means that "2+2 = 4" is a true statement, but the fact that 2+2 = 4 is NOT a consequence of the statement 2+3 = 6.

The situation we find ourselves in is usually this:

We want to know what happens if "A" happens to be true. For example:

Given that:

k + 0 = k, for all integers k

are we now in a position to explicitly evaluate 0+0?

We reason:

1. k + 0 = k, for all integers k
2. 0 is an integer (discovered by some other means, perhaps, taken as given here)
3. 0 + 0 = 0 (1, using k = 0...this is an example of substitution or instantiation-replacing a logical variable with a logical constant)

Often (1) is even given in the form:

1. There exists an integer 0, such that k + 0 = k, for all integers k

which eliminates the need for (2).

We conclude that:

k + 0 = k implies 0 + 0 = 0.

Somewhat confusingly, logical variables are often replaced by other logical variables, as we do here:

(1a) for all X in S, X + 0 = X (given)
(2a) A + B in S (given)
(3a) (A + B) + 0 = A + B (replacing the variable X with the variable A + B in (1)).

We might even replace X, with another variable dependent on X, such as X + X:

(2b) X + X in S
(3b) (X + X) + 0 = X + X

We would still conclude that:

(1a) implies (3a)

or:

(1a) implies (3b).

Usually, (3a) or (3b) is the statement we actually want to USE, it is very likely we have yet ANOTHER implication statement that will use IT as its antecedent.

Now, if (1a) turned out not to be true, all of our effort is for naught. The statements we wish to derive from it, might still be true, but we would need another way to establish that.

Axiomatic statements are usually accepted as "given" or true a priori. Occasionally this process is "reversed" in mathematics, the most famous example being the revision of the axiom of comprehension in set theory, which had to be modified because it lead to paradoxes (Russel sets).

When one is studying vector spaces, for example, there are many theorems that begin:

If $V$ is a finite-dimensional vector space, then....

Now there are many infinite-dimensional vector spaces, and such theorems may, or may not be true for these. In any case, however, the PROOFS of those theorems are not valid for the infinite-dimensional case....even if the conclusion of the original theorem is still true in the infinite-dimensional case.

Proof, in mathematics is used in two different senses, one formal and one informal.

In a formal proof, one defines an alphabet, some rules, and a logical system (usually, but not always, the classical first-order predicate logic, or something "equivalent" to it which can prove the same statements) and creates an algorithm by which the derivation of B from A could be done mechanically.

In an informal proof, the idea is to communicate enough of the skeleton of a formal proof, so that it is convincing. Implication is often buried in such phrases as:

"And we see that..."

"Now, as a consequence...."

"It suffices that..."

or the pithy:

"So...."

Instructors in any given mathematical course often differ as to the level of formalism required, and this can even vary with the SAME instructor in two different courses (rather more formalism is to be expected in a course on mathematical logic than say, in combinatorics). As such, when a question is posed on these forums, it is not always a straight-forward matter to determine what degree of sophistication and/or formality is required in our responses.

#### solakis

##### Active member
There are two options regarding validity. If we insist on deriving 2+2=4 from "If 2+3=6 then 2+2=4", i.e., if the argument is complete and there is no way to derive 2+2=4 using some other reasoning that is not shown, then the argument is not valid. It has the form "$A\to B$, $B$", and this transition does not conform to any of the inference rules. Modus Ponens could be used to derive $B$ from $A\to B$, but then the argument must also contain a proof of $A$.

The second option is that the argument is abbreviated and 2+2=4 is derived not from "If 2+3=6 then 2+2=4", but using other means that are not shown; for example, 2+2=4 is an axiom. Then the argument is valid. The premise "If 2+3=6 then 2+2=4" is simply not used.

This argument is either incomplete or invalid. Again, there are two options: either you derive "I will buy a car" essentially using the premise "If I go to London, I will buy a car", or you derive it without using that premise, using some independent means. In the first case, you also need to prove "I will go to London" in order to use Modus Ponens. In the second case, the argument is incomplete, and without extra information there is less reason to think that you will buy a car than to believe that 2 + 2 = 4.

I am afraid to say you have not given me a solid pfoof whether the two statements are valid or not.

A proof based on adefinition or a theorem or an axiom of the symbolic logic stating clearly when an argument is valid or not

#### solakis

##### Active member
I'd like to point out that statements like:

"IF 2+3 = 6, THEN 2+2 = 4"

are usually not considered within mathematical study.

This is actually a TRUE statement, because it can be proved by other means that "2+2 = 4" is a true statement, but the fact that 2+2 = 4 is NOT a consequence of the statement 2+3 = 6.

The situation we find ourselves in is usually this:

We want to know what happens if "A" happens to be true. For example:

Given that:

k + 0 = k, for all integers k

are we now in a position to explicitly evaluate 0+0?

We reason:

1. k + 0 = k, for all integers k
2. 0 is an integer (discovered by some other means, perhaps, taken as given here)
3. 0 + 0 = 0 (1, using k = 0...this is an example of substitution or instantiation-replacing a logical variable with a logical constant)

Often (1) is even given in the form:

1. There exists an integer 0, such that k + 0 = k, for all integers k

which eliminates the need for (2).

We conclude that:

k + 0 = k implies 0 + 0 = 0.

Somewhat confusingly, logical variables are often replaced by other logical variables, as we do here:

(1a) for all X in S, X + 0 = X (given)
(2a) A + B in S (given)
(3a) (A + B) + 0 = A + B (replacing the variable X with the variable A + B in (1)).

We might even replace X, with another variable dependent on X, such as X + X:

(2b) X + X in S
(3b) (X + X) + 0 = X + X

We would still conclude that:

(1a) implies (3a)

or:

(1a) implies (3b).

Usually, (3a) or (3b) is the statement we actually want to USE, it is very likely we have yet ANOTHER implication statement that will use IT as its antecedent.

Now, if (1a) turned out not to be true, all of our effort is for naught. The statements we wish to derive from it, might still be true, but we would need another way to establish that.

Axiomatic statements are usually accepted as "given" or true a priori. Occasionally this process is "reversed" in mathematics, the most famous example being the revision of the axiom of comprehension in set theory, which had to be modified because it lead to paradoxes (Russel sets).

When one is studying vector spaces, for example, there are many theorems that begin:

If $V$ is a finite-dimensional vector space, then....

Now there are many infinite-dimensional vector spaces, and such theorems may, or may not be true for these. In any case, however, the PROOFS of those theorems are not valid for the infinite-dimensional case....even if the conclusion of the original theorem is still true in the infinite-dimensional case.

Proof, in mathematics is used in two different senses, one formal and one informal.

In a formal proof, one defines an alphabet, some rules, and a logical system (usually, but not always, the classical first-order predicate logic, or something "equivalent" to it which can prove the same statements) and creates an algorithm by which the derivation of B from A could be done mechanically.

In an informal proof, the idea is to communicate enough of the skeleton of a formal proof, so that it is convincing. Implication is often buried in such phrases as:

"And we see that..."

"Now, as a consequence...."

"It suffices that..."

or the pithy:

"So...."

Instructors in any given mathematical course often differ as to the level of formalism required, and this can even vary with the SAME instructor in two different courses (rather more formalism is to be expected in a course on mathematical logic than say, in combinatorics). As such, when a question is posed on these forums, it is not always a straight-forward matter to determine what degree of sophistication and/or formality is required in our responses.

Does the above answer my questions:

1) What an algebraic argument consists of

2) What is a logical proposition ??

#### Deveno

##### Well-known member
MHB Math Scholar
Does the above answer my questions:

1) What an algebraic argument consists of

2) What is a logical proposition ??
No, I'm afraid it doesn't. I feel that #2 is easier to answer in some contexts, but there's a few things you have to define before-hand (atoms, connectives, and allowed combinations of these).

#1 is somewhat problematic: most algebraic arguments fall into the "informal" category, it is assumed before-hand that the reader is somewhat accustomed to how these things go.

An example:

Proof that the additive identity (zero) of the integers is unique:

Assume that $e \neq 0$ is another additive identity.

Then $0 = 0 + e = e$, contradiction.

Obviously, as a formal proof, this has several flaws, but it is nonetheless typical of proofs you would find in an algebra textbook; and the reasoning behind it should be clear.