Simultaneous Linear Inequalities: Yesterday and Today
aa r X i v : . [ m a t h . O C ] J u l SIMULTANEOUS LINEAR INEQUALITIES:YESTERDAY AND TODAY
S. S. KUTATELADZE
Abstract.
This is an short overview of the recent tendencies in the theory oflinear inequalities that are evoked by Boolean valued analysis. Agenda
Linear inequality implies linearity and order. When combined, the two producean ordered vector space. Each linear inequality in the simplest environment of thesort is some half-space. Simultaneity implies many instances and so leads to theintersections of half-spaces. These yield polyhedra as well as arbitrary convex sets,identifying the theory of linear inequalities with convexity.Convexity reigns in the federation of geometry, optimization, and functional anal-ysis. Convexity feeds generation, separation, calculus, and approximation. Gener-ation appears as duality; separation, as optimality; calculus, as representation; andapproximation, as stability (cp. [1]).This talk addresses the origin and the state of the art of the relevant areas with aparticular emphasis on the Farkas Lemma (cp. [2]). Our aim is to demonstrate howBoolean valued analysis may be applied to simultaneous linear inequalities withoperators.This particular theme is another illustration of the deep and powerful techniqueof “stratified validity” which is characteristic of Boolean valued analysis.2.
Founding Fathers
Linearity, inequality, and convexity stem from the remote ages (cp. [3]–[5]). How-ever, as the acclaimed pioneers who propounded these ideas and anticipated theirsignificance for the future we must rank the three polymaths,
Joseph-Louis La-grange (January 25, 1736–April 10, 1813),
Jean Baptiste Joseph Fourier (March21, 1768–May 16, 1830), and
Hermann Minkowski (June 22, 1864–January 12,1909).The translator of the famous elementary textbook of Lagrange [6] Thomas Mc-Cormack remarked:
In both research and exposition, he totally reversed the methods of his prede-cessors. They had proceeded in their exposition from special cases by a species ofinduction; his eye was always directed to the highest and most general points ofview; and it was by his suppression of details and neglect of minor, unimportant
Date : July 10, 2010.This article bases on a talk at the opening session of the International Conference “Order Anal-ysis and Related Problems of Mathematical Modeling,” Vladikavkaz, July 19–24, 2010, dedicatedto the 10th anniversary of the Vladikavkaz Scientific Center of the Russian Academy of Sciences. considerations that he swept the whole field of analysis with a generality of insightand power never excelled, adding to his originality and profundity a conciseness,elegance, and lucidity which have made him the model of mathematical writers.
The pivotal figure was Fourier. Jean-Pierre Kahane wrote in [7, pp. 83–84]:
He himself was neglected for his work on inequalities, what he called “Anal-yse ind´etermin´ee.” Darboux considered that he gave the subject an exaggeratedimportance and did not publish the papers on this question in his edition of thescientific works of Fourier. Had they been published, linear programming andconvex analysis would be included in the heritage of Fourier.
David Hilbert lamented the untimely death of Minkowski as follows: Since my student years Minkowski was my best, most dependable friend whosupported me with all the depth and loyalty that was so characteristic of him. Ourscience, which we loved above all else, brought us together; it seemed to us a gardenfull of flowers. In it, we enjoyed looking for hidden pathways and discovered manya new perspective that appealed to our sense of beauty, and when one of us showedit to the other and we marvelled over it together, our joy was complete. He wasfor me a rare gift from heaven and I must be grateful to have possessed that giftfor so long. Now death has suddenly torn him from our midst. However, whatdeath cannot take away is his noble image in our hearts and the knowledge thathis spirit in us continue to be active. Environment
Assume that X is a real vector space, Y is a Kantorovich space also known asa complete vector lattice or a Dedekind complete Riesz space. Let B := B ( Y ) bethe base of Y , i.e., the complete Boolean algebras of positive projections in Y ;and let m ( Y ) be the universal completion of Y . Denote by L ( X, Y ) the spaceof linear operators from X to Y . In case X is furnished with some Y -seminormon X , by L ( m ) ( X, Y ) we mean the space of dominated operators from X to Y . Asusual, { T ≤ } := { x ∈ X | T x ≤ } ; ker( T ) = T − (0) for T : X / / Y . Also, P ∈ Sub(
X, Y ) means that P is sublinear , while P ∈ PSub(
X, Y ) means that P is polyhedral , i.e., finitely generated. The superscript ( m ) suggests domination.4. Kantorovich’s Theorem
Find X satisfying X B BBBBBBBB A / / W X (cid:15) (cid:15) Y ( ( ∃ X ) X A = B ↔ ker( A ) ⊂ ker( B ) . ( If W is ordered by W + and A ( X ) − W + = W + − A ( X ) = W , then ( ∃ X ≥ X A = B ↔ { A ≤ } ⊂ { B ≤ } . Cp. [8]. Also see [9]. Cp. [10, p. 51]. The Alternative
Let X be a Y -seminormed real vector space, with Y a Kantorovich space. As-sume that A , . . . , A N and B belong to L ( m ) ( X, Y ) .Then one and only one of the following holds: (1) There are x ∈ X and b, b ′ ∈ B such that b ′ ≤ b and b ′ Bx > , bA x ≤ , . . . , bA N x ≤ . (2) There are positive orthomorphisms α , . . . , α N ∈ Orth( m ( Y )) + such that B = P Nk =1 α k A k . Reals: Hidden Dominance
Lemma 1.
Let X be a vector space over some subfield R of the reals R . Assumethat f and g are R -linear functionals on X ; in symbols, f, g ∈ X := L ( X, R ) .For the inclusion { g ≤ } ⊃ { f ≤ } to hold it is necessary and sufficient that there be α ∈ R + satisfying g = αf . Proof. Sufficiency is obvious.
Necessity:
The case of f = 0 is trivial. If f = 0 then there is some x ∈ X suchthat f ( x ) ∈ R and f ( x ) >
0. Denote the image f ( X ) of X under f by R . Put h := g ◦ f − , i.e. h ∈ R is the only solution for h ◦ f = g . By hypothesis, h is apositive R -linear functional on R . By the Bigard Theorem h can be extended toa positive homomorphism ¯ h : R / / R , since R − R + = R + − R = R . Each positiveautomorphism of R is multiplication by a positive real. As the sought α we maytake ¯ h (1).The proof of the lemma is complete.7. Reals: Explicit Dominance
Lemma 2.
Let X be an R -seminormed vector space over some subfield R of R .Assume that f , . . . , f N and g are bounded R -linear functionals on X ; in symbols, f , . . . , f N , g ∈ X ∗ := L ( m ) ( X, R ) .For the inclusion { g ≤ } ⊃ N \ k =1 { f k ≤ } to hold it is necessary and sufficient that there be α , . . . , α N ∈ R + satisfying g = P Nk =1 α k f k . Farkas: Explicit Dominance
Theorem 1.
Assume that A , . . . , A N and B belong to L ( m ) ( X, Y ) .The following are equivalent: (1) Given b ∈ B , the operator inequality bBx ≤ is a consequence of the simul-taneous linear operator inequalities bA x ≤ , . . . , bA N x ≤ , i.e., { bB ≤ } ⊃ { bA ≤ } ∩ · · · ∩ { bA N ≤ } . Cp. [10, p. 108].
S. S. KUTATELADZE (2)
There are positive orthomorphisms α , . . . , α N ∈ Orth( m ( Y )) such that B = N X k =1 α k A k ; i.e., B lies in the operator convex conic hull of A , . . . , A N . Boolean Modeling
Cohen’s final solution of the problem of the cardinality of the continuum withinZFC gave rise to the Boolean valued models by Scott, Solovay, and Vopˇenka. Takeuti coined the term
Boolean valued analysis for applications of the new modelsto analysis. Scott forecasted in 1969: We must ask whether there is any interest in these nonstandard models asidefrom the independence proof; that is, do they have any mathematical interest?The answer must be yes, but we cannot yet give a really good argument.
In 2009 Scott wrote: At the time, I was disappointed that no one took up my suggestion. And thenI was very surprised much later to see the work of Takeuti and his associates. Ithink the point is that people have to be trained in Functional Analysis in orderto understand these models. I think this is also obvious from your book and itsreferences. Alas, I had no students or collaborators with this kind of background,and so I was not able to generate any progress .10.
Boolean Valued Universe
Let B be a complete Boolean algebra. Given an ordinal α , put V ( B ) α := { x | ( ∃ β ∈ α ) x : dom( x ) → B & dom( x ) ⊂ V ( B ) β } . The
Boolean valued universe V ( B ) is V ( B ) := [ α ∈ On V ( B ) α , with On the class of all ordinals.The truth value [[ ϕ ]] ∈ B is assigned to each formula ϕ of ZFC relativized to V ( B ) .11. Descending and Ascending
Given ϕ , a formula of ZFC, and y , a member of VB ; put A ϕ := A ϕ ( · , y ) := { x | ϕ ( x, y ) } .The descent A ϕ ↓ of a class A ϕ is A ϕ ↓ := { t | t ∈ V ( B ) & [[ ϕ ( t, y )]] = } . If t ∈ A ϕ ↓ , then it is said that t satisfies ϕ ( · , y ) inside V ( B ) .The descent x ↓ of x ∈ V ( B ) is defined as x ↓ := { t | t ∈ V ( B ) & [[ t ∈ x ]] = } , Cp. [11]. Cp. [12]. Cp. [13]. A letter of April 29, 2009 to S. S. Kutateladze. i.e. x ↓ = A ·∈ x ↓ . The class x ↓ is a set.If x is a nonempty set inside V ( B ) then( ∃ z ∈ x ↓ )[[( ∃ t ∈ x ) ϕ ( t )]] = [[ ϕ ( z )]] . The ascent functor acts in the opposite direction.12.
The Reals Within
There is an object R inside V ( B ) modeling R , i.e.,[[ R is the reals ]] = . Let R ↓ be the descent of the carrier | R | of the algebraic system R := ( | R | , + , · , , , ≤ ) inside V ( B ) .Implement the descent of the structures on | R | to R ↓ as follows: x + y = z ↔ [[ x + y = z ]] = ; xy = z ↔ [[ xy = z ]] = ; x ≤ y ↔ [[ x ≤ y ]] = ; λx = y ↔ [[ λ ∧ x = y ]] = ( x, y, z ∈ R ↓ , λ ∈ R ) . Gordon Theorem. R ↓ with the descended structures is a universally completevector lattice with base B ( R ↓ ) isomorphic to B .13. Proof of Theorem 1 (2) / / (1): If B = P Nk =1 α k A k for some positive α , . . . , α N in Orth( m ( Y )) while bA k x ≤ b ∈ B and x ∈ X , then bBx = b N X k =1 α k A k x = N X k =1 α k bA k x ≤ m ( Y ).(1) / / (2): Consider the separated Boolean valued universe V ( B ) over the base B of Y . By the Gordon Theorem the ascent Y ↑ of Y is R , the reals inside V ( B ) .Using the canonical embedding, we see that X ∧ is an R -seminormed vector spaceover the standard name R ∧ of the reals R .Moreover, R ∧ is a subfield and sublattice of R = Y ↑ inside V ( B ) .Put f k := A k ↑ for all k := 1 , . . . , N and g := B ↑ . Clearly, all f , . . . , f N , g belongto ( X ∧ ) ∗ inside VB .Define the finite sequence f : { , . . . , N } ∧ / / ( X ∧ ) ∗ as the ascent of ( f , . . . , f N ). In other words, the truth values are as follows:[[ f k ∧ ( x ∧ ) = A k x ]] = , [[ g ( x ∧ ) = Bx ]] = for all x ∈ X and k := 1 , . . . , N .Put b := [[ A x ≤ ∧ ]] ∧ · · · ∧ [[ A N x ≤ ∧ ]] . Then bA k x ≤ k := 1 , . . . , N and bBx ≤ Cp. [11, p. 349].
S. S. KUTATELADZE
Therefore, [[ A x ≤ ∧ ]] ∧ · · · ∧ [[ A N x ≤ ∧ ]] ≤ [[ Bx ≤ ∧ ]] . In other words, [[( ∀ k := 1 ∧ , . . . , N ∧ ) f k ( x ∧ ) ≤ ∧ ]]= ^ k :=1 ,...,N [[ f k ∧ ( x ∧ ) ≤ ∧ ]] ≤ [[ g ( x ∧ ) ≤ ∧ ]] . By Lemma 2 inside V ( B ) and the maximum principle of Boolean valued analysis,there is a finite sequence α : { ∧ , . . . , N ∧ } / / R + inside V ( B ) satisfying[[( ∀ x ∈ X ∧ ) g ( x ) = N ∧ X k =1 ∧ α ( k ) f k ( x )]] = . Put α k := α ( k ∧ ) ∈ R + ↓ for k := 1 , . . . , N . Multiplication by an element in R ↓ is an orthomorphism of m ( Y ). Moreover, B = N X k =1 α k A k , which completes the proof.14. Counterexample: No Dominance
Lemma 1, describing the consequences of a single inequality, does not restrictthe class of functionals under consideration.The analogous version of the Farkas Lemma simply fails for two simultaneousinequalities in general.The inclusion { f = 0 } ⊂ { g ≤ } equivalent to the inclusion { f = 0 } ⊂ { g = 0 } does not imply that f and g are proportional in the case of an arbitrary subfieldof R . It suffices to look at R over the rationals Q , take some discontinuous Q -linearfunctional on Q and the identity automorphism of Q .15. Reconstruction: No Dominance
Theorem 2.
Take A and B in L ( X, Y ) . The following are equivalent: (1) ( ∃ α ∈ Orth( m ( Y ))) B = αA ; (2) There is a projection κ ∈ B such that { κ bB ≤ } ⊃ { κ bA ≤ } ; {¬ κ bB ≤ } ⊃ {¬ κ bA ≥ } for all b ∈ B . Proof.
Boolean valued analysis reduces the claim to the scalar case. ApplyingLemma 1 twice and writing down the truth values, complete the proof. As usual, ¬ κ := − κ . Interval Operators
Let X be a vector lattice. An interval operator T from X to Y is an orderinterval [ T , T ] in L ( r ) ( X, Y ), with T ≤ T . The interval equation B = X A hasa weak interval solution provided that ( ∃ X )( ∃ A ∈ A )( ∃ B ∈ B ) B = X A .Given an interval operator T and x ∈ X , put P T ( x ) = T x + − T x − . Call T adapted in case T − T is the sum of finitely many disjoint addends.Put ∼ ( x ) := − x for all x ∈ X .17. Interval Equations
Theorem 3.
Let X be a vector lattice, and let Y be a Kantorovich space.Assume that A , . . . , A N are adapted interval operators and B is an arbitraryinterval operator in the space of order bounded operators L ( r ) ( X, Y ) .The following are equivalent: (1) The interval equation B = N X k =1 α k A k has a weak interval solution α , . . . , α N ∈ Orth( Y ) + . (2) For all b ∈ B we have { b B ≥ } ⊃ { b A ∼ ≤ } ∩ · · · ∩ { b A ∼ N ≤ } , where A ∼ k := P A k ◦ ∼ for k := 1 , . . . , N and B := P B . Inhomogeneous Inequalities
Theorem 4.
Let X be a Y -seminormed real vector space, with Y a Kantorovichspace. Assume given some dominated operators A , . . . , A N , B ∈ L ( m ) ( X, Y ) andelements u , . . . , u N , v ∈ Y . The following are equivalent: (1) For all b ∈ B the inhomogeneous operator inequality bBx ≤ bv is a conse-quence of the consistent simultaneous inhomogeneous operator inequalities bA x ≤ bu , . . . , bA N x ≤ bu N , i.e., { bB ≤ bv } ⊃ { bA ≤ bu } ∩ · · · ∩ { bA N ≤ bu N } . (2) There are positive orthomorphisms α , . . . , α N ∈ Orth( m ( Y )) satisfying B = N X k =1 α k A k ; v ≥ N X k =1 α k u k . Cp. [15].
S. S. KUTATELADZE
Inhomogeneous Matrix Inequalities
Theorem 5. Let X be a Y -seminormed real vector space, with Y a Kan-torovich space. Assume that A ∈ L ( m ) ( X, Y s ) , B ∈ L ( m ) ( X, Y t ) , u ∈ Y s , and v ∈ Y t , where s and t are some naturals.The following are equivalent: (1) For all b ∈ B the inhomogeneous operator inequality bBx ≤ bv is a conse-quence of the consistent inhomogeneous inequality bAx ≤ bu , i.e., { bB ≤ bv } ⊃{ bA ≤ bu } . (2) There is some s × t matrix with entries positive orthomorphisms of m ( Y ) suchthat B = X A and X u ≤ v for the corresponding linear operator X ∈ L + ( Y s , Y t ) . Complex Scalars
Theorem 6.
Let X be a Y -seminormed complex vector space, with Y a Kan-torovich space. Assume given some u , . . . , u N , v ∈ Y and dominated operators A , . . . , A N , B ∈ L ( m ) ( X, Y C ) from X into the complexification Y C := Y ⊗ iY of Y . Assume further that the simultaneous inhomogeneous inequalities | A x | ≤ u , . . . , | A N x | ≤ u N are consistent. Then the following are equivalent: (1) { b | B ( · ) | ≤ bv } ⊃ { b | A ( · ) | ≤ bu } ∩ · · · ∩ { b | A N ( · ) | ≤ bu N } for all b ∈ B . (2) There are complex orthomorphisms c , . . . , c N ∈ Orth( m ( Y ) C ) satisfying B = N X k =1 c k A k ; v ≥ N X k =1 | c k | u k . Inhomogeneous Sublinear Inequalities
Lemma 3.
Let X be a real vector space. Assume that p , . . . , p N ∈ PSub( X ) :=PSub( X, R ) and p ∈ Sub( X ) . Assume further that v, u , . . . , u N ∈ R make consis-tent the simultaneous sublinear inequalities p k ( x ) ≤ u k , with k := 1 , . . . , N .The following are equivalent: (1) { p ≥ v } ⊃ N T k =1 { p k ≤ u k } ;(2) There are α , . . . , α N ∈ R + satisfying ( ∀ x ∈ X ) p ( x ) + N X k =1 α k p k ( x ) ≥ , N X k =1 α k u k ≤ − v. Proof. (2) / / (1): If x is a solution to the simultaneous inhomogeneous in-equalities p k ( x ) ≤ u k with k := 1 , . . . , N , then0 ≤ p ( x ) + N X k =1 α k p k ( x ) ≤ p ( x ) + N X k =1 α k u k ( x ) ≤ p ( x ) − v. Cp. [14]. Cp. [3, p. 338]. (1) / / (2): Given ( x, t ) ∈ X × R , put ¯ p k ( x, t ) := p k ( x ) − tu k , ¯ p ( x, t ) := p ( x ) − tv and τ ( x, t ) := − t . Clearly, τ, ¯ p , . . . , ¯ p N ∈ PSub( X × R ) and ¯ p ∈ Sub( X × R ). Take( x, t ) ∈ { τ ≤ } ∩ N \ k =1 { ¯ p k ≤ } . If, moreover, t >
0; then u k ≥ p k ( x/t ) for k := 1 , . . . , N and so p ( x/t ) ≤ v byhypothesis. In other words ( x, t ) ∈ { ¯ p ≤ } . If t = 0 then take some solution ¯ x ofthe simultaneous inhomogeneous polyhedral inequalities under study.Since x ∈ K := T Nk =1 { p k ≤ } ; therefore, p k (¯ x + x ) ≤ p ( x ) + p k ( x ) ≤ u k for all k := 1 , . . . , N . Hence, p (¯ x + x ) ≥ v by hypothesis. So the sublinear functional p isbounded below on the cone K . Consequently, p assumes only positive values on K .In other words, ( x, ∈ { ¯ p ≤ } . Thus { ¯ p ≥ } ⊃ N \ k =1 { ¯ p k ≤ } and by Lemma 2.2. of [1] there are positive reals α , . . . , α N , β such that for all( x, t ) ∈ X × R we have ¯ g ( x ) + βτ ( x ) + N X k =1 α k ¯ p k ( x ) ≥ . Clearly, the so-obtained parameters α , . . . , α N are what we sought for. The proofof the lemma is complete. Corollary.
Let X be an R -seminormed complex vector space. Given are u , . . . , u N , v ∈ Y and bounded complex linear functionals f , . . . , f N , f ∈ X ∗ .Assume that consistent are the simultaneous inhomogeneous inequalities | f ( x ) | ≤ u , . . . , | f N ( x ) | ≤ u N . Then the following are equivalent: (1) The inequality | g ( x ) | ≤ v is a consequence of the simultaneous inequalities | f ( x ) | ≤ u , . . . , | f N ( x ) | ≤ u N , i.e. {| g ( · ) | ≤ v } ⊃ {| f ( · ) | ≤ u } ∪ · · · ∪ {| f N ( · ) | ≤ u N } ;(2) There are c , . . . , c N ∈ C such that g = N X k =1 c k f k , v ≥ N X k =1 | c k | u k . Proof. (2) / / (1): If x ∈ T Nk =1 {| f k ( · ) | ≤ u k } then | g ( x ) | = (cid:12)(cid:12) N X k =1 c k f k ( x ) (cid:12)(cid:12) ≤ N X k =1 | c k f k ( x ) | ≤ N X k =1 | c k | u k ≤ v. (1) / / (2): Consider the realification X R of X and the sublinear functionals p ( x ) := − Re g ( x ) and p k ( x ) := | f k ( x ) | , where k := 1 , . . . , N and x ∈ X R . Clearly,Lemma 3 applies and { p ≥ − v } ⊃ N \ k =1 { p k ≤ u k } . Hence, there are positive reals α , . . . , α N satisfying( ∀ x ∈ X R ) − Re g ( x ) + N X k =1 α k | f k ( x ) | ≥ N X k =1 α k u k ≤ v. By subdifferential calculus there are complexes θ k , | θ k | = 1 , k := 1 , . . . , N , such that g = P α k θ k f k . Put c k := α k θ k . Obviously, P Nk =1 | c k | u k = P Nk =1 α k | θ k | u k ≤ v .The proof of the corollary is complete. Remark.
Theorem 6 (which is Theorem 3.1 in [20] supplied with a slightlydubious proof) is a Boolean valued interpretation of the Corollary.
Theorem 7.
Let X be a Y -seminormed real vector space, with Y a Kantorovichspace. Given are some dominated polyhedral sublinear operators P , . . . , P N ∈ PSub ( m ) ( X, Y ) and a dominated sublinear operator P ∈ Sub ( m ) ( X, Y ) . Assumefurther that u , . . . , u N , v ∈ Y make consistent the simultaneous inhomogeneousinequalities P ( x ) ≤ u , . . . , P N ( x ) ≤ u N , P ( x ) ≥ v .The following are equivalent: (1) For all b ∈ B the inhomogeneous sublinear operator inequality bP ( x ) ≥ v isa consequence of the simultaneous inhomogeneous sublinear operator inequalities bP ( x ) ≤ u , . . . , bP N ( x ) ≤ u N , i.e., { bP ≥ v } ⊃ { bP ≤ u } ∩ · · · ∩ { bP N ≤ u N } ;(2) There are positive α , . . . , α N ∈ Orth( m ( Y )) satisfying ( ∀ x ∈ X ) P ( x ) + N X k =1 α k P k ( x ) ≥ , N X k =1 α k u k ≤ − v. Lagrange’s Principle
The finite value of the constrained problem P ( x ) ≤ u , . . . , P N ( x ) ≤ u N , P ( x ) / / infis the value of the unconstrained problem for an appropriate Lagrangian withoutany constraint qualification other that polyhedrality.The Slater condition allows us to eliminate polyhedrality as well as consideringa unique target space. This is available in a practically unrestricted generality. About the new trends relevant to the Farkas Lemma see [16]–[20].23.
Freedom and Inequality
Convexity is the theory of linear inequalities in disguise.Abstraction is the freedom of generalization. Freedom is the loftiest ideal andidea of man, but it is demanding, limited, and vexing. So is abstraction. So are itsinstances in convexity, hence, in simultaneous inequalities.We definitely feel truth, but we cannot define truth properly. That is what AlfredTarski explained to us in the 1930s. We pursue truth by way of proof, as wittilyphrased by Saunders Mac Lane. Mathematics becomes logic.The freedom of set theory empowered us with the Boolean valued models yieldinga lot of surprising and unforeseen visualizations of the ingredients of mathematics. Cp. [10]. Cp. [21]. Cp. [22]. Many promising opportunities are open to modeling the powerful habits of reason-ing and verification.Logic organizes and orders our ways of thinking, manumitting us from conser-vatism in choosing the objects and methods of research. Logic of today is a fineinstrument and institution of mathematical freedom. Logic liberates mathematicsby model theory.Model theory evaluates and counts truth and proof. The chase of truth notonly leads us close to the truth we pursue but also enables us to nearly catch upwith many other instances of truth which we were not aware nor even foresaw atthe start of the rally pursuit. That is what we have learned from Boolean valuedanalysis.Freedom presumes liberty and equality. Inequality paves way to freedom.
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