# Discussion of 'Valid method of proof'.

dx said and i quote:A proof is a proof: No.
But you said $$\forall$$ x (x=x) So therefore, a proof is a proof! :-)

Do you mean that from now on the axiom in real Nos (For all x and y x+y=y+x) we should call it a THEOREM?
when you have lets say two sta/nts P and Q which are logicaly equivalent (P===>Q),if you start from P to prove Q P is called axiom and Q theorem,and since are equivalent when we start from Q to prove P Q is called axiom and P THEOREM
Now the axiom of choice,the Well ordering theorem,Zorn's Lemma are all equivalent and depending where you start from you call it an axiom and the rest thoerms

matt grime
Homework Helper
Yes, for the umpteenth time it is conventionally taken that an axiom is *trivially* a theorem - it can be deduced from itself since (X=>X) is always true. If you don't like that then you're just arguing against the convention that the mathematical community has adopted, which is going to be a fruitless endeavour.

And the point is irrelevant anyway since I've given you plenty of other examples of showing how to deduce a (true) non-vacuous statement from a false one. That I've been able to do so doesn't prove that this non-vacuous statement is true, obviously, which is what you still appear to think we're telling you.

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A THING that implies itself ((X==>X)<===>(X OR notX)) IT does not necesseraly mean that it looses its name
Now if the mathematical community decides to call the axioms theorems ,and since X IMPLIES X and not Y the thoerems axioms i will be the last person on earth to say otherwise
But inform the mathematical community to change the books too
At this point please let me take arest because i have given the Forum couple of problems and since nobody has responded yet i must at least give a sample solution

Let me now prove IxI<y<===> --y<x<y the way i said
NOTE (L) will mean law of logic ,(T) A theorem,(D) a definition,(A) an axiom (x>=y
) x> or =toy
So
1) IxI<y an assumption
2) for all x x>=0 or x< 0 (A)
3) x>=0 or x<0 from line2 and using Universal elimination (L)
4) x>=0 A hypothesis
5) for all x x>=0 ------> IxI=x (D) in apsolute values
6) x>=0-------> IxI=x from 5 and Univ. elimin. (L)
7) IxI=x from 4 and 6 and using M.Ponens (L)
8) x<y By substituting 7 into 1 (L)
9) for all a,b,c a<=b and b<c -----> a<c (T) or a result coming out from iniqalities
10) 0<=x and x<y-----> 0<y from 9 and Univ.Elim.where we put a=0,b=x,c=y
11) 0<=x and x<y from 4 and 8 and using Conjuction Introduction (L)
12) 0<y from 10 and 11 and using M.Ponens (L)
13) for all a<b<----->-a>-b (T)
14) 0<y<-----> 0>-y from 13 and Univ. Elim. where a=0,b=y
15) -y<0 from 12 and 14 and using M.Ponens
16) for all a<b and b<=c------> a<c (T)
17) -y<0 and 0<=x------> -y<x from 16 and Univ. Elim.where we put a=-y,b=0,c=x
18) -y<0 and 0<=x from15 and 4 and using conjuction introduction (L)
19 -y<x from 17 and 18 and using M.Ponens
20) -y<x and x<y ( -y<x<y) from 8 and 19 and conjuction introduction (L)
21) x>=0-----> -y<x<y from steps 4 to 21 and using the rule of conditional proof (L)
Now in a similar way and using the definition of apsolute values x<0----> IxI=-x we will come to the result x<0-----> -y<x<y ( 1a)
22) -y<x<y from 2, (1a),21 and Disjuction Elimination (L)
23) IXI<y-----> -y<x<y from steps 1 to 23 and using the rule of conditinal proof
the converse i,e -y<x<y----> IxI<y can be done in a similar way

The laws of logic used in this proof are:
1) M.Ponens ( P--->Q and P)====>Q (Where a single arrow means an implication and the double one a LOGICAL IMPLICATION)
2) Conjuction introduction P,Q =====> P and Q
3)Disjuction Elimination ( P---->S and Q---->S and P or Q)=====>S
4)The law
of Universal Elimination where if aproperty holds for a set it will hold for one of its members
5)The law of substitution
6)The law of conditional proof in some books is called the deduction theorem an it is considered some times as ametatheorem
The theorems the definition and the Trichotomy Law are well known to every one they are highschool stuff

This type of proof leaves no room for imagining things in mathematics or arguing for ever and ever or a teacher in every level to pass wrong proofs to the students because he likes to.For people involved in hard core analysis where a lot of quantification is involved that machinery can solve most of your problems. Of course it takes a little pactise but it is worth while

It is simply the case that (F=>T) is T, i.e. false implies true is true.
Hi Matt, I have a subtle disagreement on using the explanation (F=>T) is T involving the material conditional =>. The statement that theorem A can be deduced from premises S is written as $S\vdash A$, or in some other way than a material conditional. In other words, saying that A is a consequence of B is not the same as saying A implies B.

Here is the proof that with one false statement as premise one can derive any conclusion.

Mc = I'm posting from the moon. (This is false)
(x) ~Mx = No one can post from the moon. (A true fact)
B = anything you want!

1p. Mc
2p. (x) ~Mx
---------
2. Mc v B 1, (valid logical addition, if Mc is true then Mc v B is true)
3. ~Mc 2, (universal instantiation of a specific fact)
4. B 2,3 Disjunctive syllogism

And now i will quote couple of proofs from different books and if they are right or wrong how can we establish that
So
proof No 1
Let a' mean a to the square
IxI<y<===> IxI'<y' <===> x'-y'<0 <===> (x-y).(x+y)<0 <===> { x-y<0}
{x+y>0} or
{x-y>0}
{x+y<o}
If {x-y<0} {x<y}
{x+y>0} <===> {x>-y} <====> -y<x<y
If {x-y>0} {x>y}
{x+y<0} <====> {x<-y} not true

{x+y<0} <====>

I am sorry that did not come out all right let me try again

IxI<y <===> IxI'<y' <===> x'<y' <===> x'-y'<0 <====> [(x-y<0 and x+y>0) or(x-y>0 andx+y<0)]
If (x-y<0 and x+y>0) <===> -y<x<y
If (x-y>0 and x+y<0) <===> x>y and x<-y not true

NOTE AGAIN a' means a to the square. I wonder is that proof valid and if yes or no why
NOTE again this is the proof of IxI<y <====> -y<x<y

Hurkyl
Staff Emeritus
Gold Member
Hi Matt, I have a subtle disagreement on using the explanation (F=>T) is T involving the material conditional =>. The statement that theorem A can be deduced from premises S is written as $S\vdash A$, or in some other way than a material conditional. In other words, saying that A is a consequence of B is not the same as saying A implies B.
Of course, the wonderful properties of first-order Boolean logic render all such distinctions essentially irrelevant.

T HE FOUNDER of the two values logic T,F is and will remain for ever THE GREAT ΑΡΙΣΤΟΤΕΛΗΣ. So the logic is Aristotelian Boolean is the algebra i mean Boolean algebra

Here is another proof from a book:
For x>=o we have:
(IxI<y and x>=0) <====> (x<y and x>=0) <====> 0<=x<y (1)
For x<0 we have:
(IxI<y and x<0) <=====> (-x<y and x<0) <====> (x>-y and x<0) <===> -y<x<0 (2)
Hence from (1) and (2) we have that IxI<y is true for those x for which -y<x<y .Hence the eqiuvelance IxI<y holds

Tomorrow i will write a professor's proof on question 3 and you decide whether is right or wrong

Here is another proof from a book:
For x>=o we have:
(IxI<y and x>=0) <====> (x<y and x>=0) <====> 0<=x<y (1)
For x<0 we have:
(IxI<y and x<0) <=====> (-x<y and x<0) <====> (x>-y and x<0) <===> -y<x<0 (2)
Hence from (1) and (2) we have that IxI<y is true for those x for which -y<x<y .Hence the eqiuvelance IxI<y holds

Perhaps before giving aprofessor's proof in question 3 i think i better give a step wise proof of atheorem that is closer to the axioms of the real Nos
So let us prove : For all xεR 0x=0
1) for all x, 1.x=x An axiom on real Nos2
2) 1.x=x from 1 and using Univ. Elim.
3) for all x,y,z (y+z).x= yx +zx An axiom on real Nos (distributive property)
4) (0+1).x= 0x+1x from 3and using Univ.Elim. where we put y=0,z=1,x=x
5) for all x,0+x=x An axiom in real Nos
6) 0+1=1 from 5 and using Univ.Elim.where we put x=1
7) 1x= 0x+1x by substituting 6 into 4
8) x=0x+x by substituting 1 into 7
9) for all a,b,c a=b ----> a+c=b+c An axiom in equality
10) x=ox+x -------> x+(-x)= (0x+x)+(-x) from 9 and using Univ.Elim.where we put a=x,b= 0x+x,c=(-x)
11) x+(-x)= (0x+x)+(-x) from 8 and 10 and using M.Ponens
12) for all a,b,c (a+b)+c= a+(b+c) An axiom in real Nos
13) (0x+x)+(-x) =0x +(x+(-x)) from 12 and using Univ,Elim. where we put a=0x,b=x,c=(-x)
14) x+(-x) =0x +(x+(-x)) by substituting 13 into 11
15) for all x, x+(-x) =0 An axiom in real Nos
16) x+(-x)=0 from 15 and using Univ.Elim.
17) 0 = 0x +0 by substituting 16 into 14
18) 0x + 0 = 0x from 5 and using Univ.Elim. where we put x=0x
19) 0 = 0x by substituting 18 into 17

Again here the laws used are:
1)Universal Elimination
2)Law of substitution
3)M.Ponens
And the axioms on real Nos:
1)For all, x 1x=x
2)For all, x,y,z (y+z)x= yx+zx
3)For all, x 0+x =x
4)For all, a,b,c a=b -----> a+c=b+c An axiom in equality
5)For all, a,b,c (a+b)+c = a+(b+c)
6)For all, x x+(-x) = 0

One may wonders O.K all that but how do we start to prove the above identity ( 0x=x).
In proving identities we start from
1)either L.H.S and end on R.H.S
2)OR R.H.S and end on L.H.S
3) or both sides and end on a common algebraic expression and in our case:
0x = 0x +0= 0x +(x+(-x)) = (0x + x)+(-x) = (0+1)x +(-x) = 1x +(-x)= x+(-x) =0
Another type of proof that have been mentioned previously is the double implications one i.e 0x=0<===> ox+x= o+x<====> (0+I)x=0+x<===>1x= 0+x<===> x=x
This type of proof assumes 0x=0 valid ( athing that we want to prove ) and with double implications end up on a valid statement

Let me now try to prove a well known limit which in ordinary mathematical books is proved within 5-6 lines. The limit is: lim(1/n)=0 as n-->oo (infinity)
According to the definition of a limit of a sequence we must prove that:
for all e>0 there exists an No ε N (natural numbers) such that for all n, n>=No then
|1/n-0|<e
1) e>0 assumption
2) for all, a (a>0 --> 1/a>0) theorem in inequalities
3) e>0 --> 1/e >0 from 2 and Univ.Elim.
4) 1/e>0 from 1 and 3 and M.Ponens
5) for all h (h>0 --> there exists No (No ε N and No>h)) theorem in real Nos known as Archimedean property
6) 1/e> 0 --> there exists No (No ε N and No>1/e) from 5 and Univ.Elim. where we put h=1/e
7) There exists No (No ε N and No>1/e) from 4 and 6 and M.Ponens
8) No ε N and No>1/e hypothesis for the choose rule
9) No ε N from 8 and conjuction elimination
10) 1/e<No from 8 and conjuction elimination
10a) n>=No hypothesis
11) for all k, (k ε N --> k>0) theorem in natural Nos
12) No ε Ν --> Νο > 0 from 11 and Univ.Elim. where we put k=No
13) No>0 from 9 and 12 and M.Ponens
14) For all a,b [0<a-->(a<=b<-->1/b<=1/a)] theorem in inequalities
15) 0 <No --> (No<=n <--> 1/n<=1/No) from 14 and Univ. Elim. where we put a=No and b=n

16) No <= n <--> 1/n <= 1/No from 15 and 13 and M. Ponens
17) 1/n <= 1/No from 10a and 16 and M.Ponens
18) for all a,b [0<a-->(a<b<-->1/b<1/a)] theorem in inequalities
19) 0 < 1/e --> (1/e<No <--> 1/No<e) from 20 and Univ. Elim. where we put a=1/e and b=No
20) 1/e<No <--> 1/No<e from 4 and 19 and M.Ponens
21) 1/No<e from 10 and 20 and M.Ponens
22) for all,a,b,c (a<=b and b<c ------> a<c) a theorem in real Nos
23) 1/n<= 1/No and 1/No<e ------> 1/n<e from 22 and Univ.Elim. where we put a=1/n,b= 1/No and c=e
24) 1/n<= 1/No and 1/No<e from 17 and 21 and conjuction introduction
25) 1/n<e from 23 and 24 and M.Ponens
26) for all a,b,c (a<b and b<=c-------> a<c) a theorem in inequalities
27) 0<No and No<=n -------->0<n from 26 and Univ.Elim. where we put a=0,b=No and c=n
28) 0<No and No<=n from 10a and 13 and conjuction introduction
29) 0<n from 27 and 28 and M.Ponens
30) 0<n-------> 0< 1/n from 2 and Univ.Elim. where we put a=n
31) 0<1/n from 29 and 30 and M.Ponens
32) for all x, (0<x -------> IxI=x) definition in absolute values
33) 0<1/n -------> I1/nI= I1/n-0I= 1/n from 32 and Univ.Elim. where we put x=1/n
34) I1/nI=I1/n-0I=1/n from 31 and 33 and M.Ponens
35) I1/nI<e by substituting 34 into 25
36) n>=No----------> I1/nI<e from steps 10a to 35 by using the conditional proof rule
37) for all n( n>=No -------------> I1/nI <e) from 36 and Univ.Introduction
38) Noε Ν and for all n( n>=No ------------> I1/nI<e) from 9 and 37 and conjuction introduction
39) there exists No ( No ε Ν and for all n( n>=No----------> I1/nI<e)) from 38 and existential introduction
40) there exists No ( No ε Ν and for all n( n>=No----------> I1/nI<e)) from steps 8 to 39 where according to the choose rule of logic we can discharge the hypothesis of step 8 and be left with the result of 39
41) e> 0--------> there exists No ( No ε Ν and for all n( n>=No----------> I1/nI<e)) from steps 1 to 40 and using the rule of conditional proof
42) for all e [ e>0--------> there exists No ( No ε Ν and for all n( n>=No----------> I1/nI<e)) from 41 and Univ. Introduction
HENCE WE HAVE PROVED lim 1/n =0 as n-----------> 00(infinity)

Well don't you use alot of unproven lemmas (i.e. 2,5,11,14,...)?

From step 10a the proof can be shortened considerably as follows:
Since NoεΝ ===> No>0 and since n>=No ===> 1/n<= 1/No ====> 1/n<e since 1/No<e , and also 1/n> 0 ====> I1/nI= 1/n since n>0. Provided we have Good command in inequalities a must in mathematical analysis.
Hence we can go straight to step 35.
Very important in analysis is,WHENEVER we start a hypothesis or assumption we must close it somewhere along the proof.
This happens particularly with proofs where we have a lot of quantification

Further more in this proof central theorem used is the Archimidean property which we can prove using the Completeness axiom in Real Nos

Do you know how many unproven theorems, lemmas, propositions,corollaries ordinary mathematical analysis books use in their proofs that they do not even mention???
WELL here at least are mentioned EXPLICITLY ,AND if you like you can check trier validity.
Besides this is the idea of the whole process ,everything to be brought up ,logic laws,theorems e.t.c so that there can be no doubt whatsoever of their validity.
Now if you can find another proof with less steps and fewer unproven theorems PLEASE DO IT

Okay, if you didn't mean "naive", what in the world is a "nabeve doctrine"?
When he wrote "your nabeve doctrine," he meant to write, "your above doctrine."

DJ

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