solakis said:
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 : $$A+B\neq 0$$ if A+B=A and $$A\neq 0$$
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).