- #26

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But you said [tex]\forall[/tex] x (x=x) So therefore, a proofdx said and i quote:A proof is a proof: No.

**is**a proof! :-)

- Thread starter LAVRANOS
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- #26

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But you said [tex]\forall[/tex] x (x=x) So therefore, a proofdx said and i quote:A proof is a proof: No.

- #27

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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

- #28

matt grime

Science Advisor

Homework Helper

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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.

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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- #29

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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

- #30

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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

- #31

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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

- #32

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- #33

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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 [itex] S\vdash A[/itex], 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.It is simply the case that (F=>T) is T, i.e. false implies true is true.

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

- #34

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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} <====>

- #35

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

- #36

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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

- #37

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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Of course, the wonderful properties of first-order Boolean logic render all such distinctions essentially irrelevant.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 [itex] S\vdash A[/itex], 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.

- #38

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- #39

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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

- #40

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

- #41

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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

- #42

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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

- #43

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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

- #44

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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

- #45

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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)

- #46

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

- #47

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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

- #48

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- #49

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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

- #50

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

DJ

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