DRAFT Distributivity in residuated structures Wesley Fussner - PowerPoint PPT Presentation

DRAFT Distributivity in residuated structures Wesley Fussner University of Denver Department of Mathematics (Joint work with P. Jipsen) BLAST 2018 Denver, CO, USA 6 August 2018 1 / 21

DRAFT The setting Definition: A residuated binar is an expansion of a lattice A by binary operations · , \ , / that satisfy the law of residuation, 2 / 21

DRAFT The setting Definition: A residuated binar is an expansion of a lattice A by binary operations · , \ , / that satisfy the law of residuation, i.e., for all x , y , z ∈ A , y ≤ x \ z ⇐ ⇒ x · y ≤ z ⇐ ⇒ x ≤ z / y . 2 / 21

DRAFT The setting Definition: A residuated binar is an expansion of a lattice A by binary operations · , \ , / that satisfy the law of residuation, i.e., for all x , y , z ∈ A , y ≤ x \ z ⇐ ⇒ x · y ≤ z ⇐ ⇒ x ≤ z / y . Note: Residuated lattices are residuated binars for which · is associative and commutative. 2 / 21

DRAFT Residuation and distributivity One may think of the law of residuation as a “finite approximation” of an infinite distributive property. 3 / 21

DRAFT Residuation and distributivity One may think of the law of residuation as a “finite approximation” of an infinite distributive property. If A = ( A , ∧ , ∨ , · , \ , / ) is a residuated binar, then multiplication preserves existing joins in each argument. 3 / 21

DRAFT Residuation and distributivity One may think of the law of residuation as a “finite approximation” of an infinite distributive property. If A = ( A , ∧ , ∨ , · , \ , / ) is a residuated binar, then multiplication preserves existing joins in each argument. In other words, if X , Y ⊆ A and � X and � Y exist, then � � � X · Y = { xy : x ∈ X , y ∈ Y } . 3 / 21

DRAFT Residuation and distributivity (cont.) The division operations \ and / also satisfy strong distributive properties. 4 / 21

DRAFT Residuation and distributivity (cont.) The division operations \ and / also satisfy strong distributive properties. Divisions preserve all existing meets in the numerator, and convert all existing joins in the denominator to meets. 4 / 21

DRAFT Residuation and distributivity (cont.) The division operations \ and / also satisfy strong distributive properties. Divisions preserve all existing meets in the numerator, and convert all existing joins in the denominator to meets. In other words, if X , Y ⊆ A and � X , � Y exist, then for any z ∈ A each of � x ∈ X x \ z , � x ∈ X z / x , � y ∈ Y z \ y , and � y ∈ Y y / z exists and � � � � z \ ( Y ) = x \ y , ( Y ) / z = y / z . y ∈ Y y ∈ Y � � � � ( X ) \ z = x \ z , z / ( X ) = z / x . x ∈ X x ∈ X 4 / 21

DRAFT Residuation and distributivity (cont.) There is a partial converse to the above. 5 / 21

DRAFT Residuation and distributivity (cont.) There is a partial converse to the above. Proposition: Let A be a complete lattice expanded by a binary operation · . Then · is residuated if it distributes over arbitrary joins in each coordinate. 5 / 21

DRAFT Residuation and distributivity (cont.) There is a partial converse to the above. Proposition: Let A be a complete lattice expanded by a binary operation · . Then · is residuated if it distributes over arbitrary joins in each coordinate. In particular, if · is a binary operation on a finite lattice then · is residuated if it satisfies x · ( y ∨ z ) = ( x · y ) ∨ ( x · z ) and ( x ∨ y ) · z = ( x · z ) ∨ ( y · z ). 5 / 21

DRAFT Residuation and distributivity (cont.) In more finitary terms, we have: 6 / 21

DRAFT Residuation and distributivity (cont.) In more finitary terms, we have: Proposition: Let A be a residuated binar. Then A satisfies the following. ( ·∨ ) x ( y ∨ z ) = xy ∨ xz . ( ∨· ) ( x ∨ y ) z = xz ∨ yz . ( \∧ ) x \ ( y ∧ z ) = x \ y ∧ x \ z . ( ∧ / ) ( x ∧ y ) / z = x / z ∧ y / z . ( / ∨ ) x / ( y ∨ z ) = x / y ∧ x / z . ( ∨\ ) ( x ∨ y ) \ z = x \ z ∧ y \ z . 6 / 21

DRAFT Nontrivial distributive laws Consider the following nontrivial distributive laws: 7 / 21

DRAFT Nontrivial distributive laws Consider the following nontrivial distributive laws: x ( y ∧ z ) = xy ∧ xz ( ·∧ ) ( x ∧ y ) z = xz ∧ yz ( ∧· ) x \ ( y ∨ z ) = x \ y ∨ x \ z ( \∨ ) ( x ∨ y ) / z = x / z ∨ y / z ( ∨ / ) ( x ∧ y ) \ z = x \ z ∨ y \ z ( ∧\ ) x / ( y ∧ z ) = x / y ∨ x / z ( / ∧ ) 7 / 21

DRAFT Nontrivial distributive laws Consider the following nontrivial distributive laws: x ( y ∧ z ) = xy ∧ xz ( ·∧ ) ( x ∧ y ) z = xz ∧ yz ( ∧· ) x \ ( y ∨ z ) = x \ y ∨ x \ z ( \∨ ) ( x ∨ y ) / z = x / z ∨ y / z ( ∨ / ) ( x ∧ y ) \ z = x \ z ∨ y \ z ( ∧\ ) x / ( y ∧ z ) = x / y ∨ x / z ( / ∧ ) The main purpose of this work is to understand the poset of subvarieties axiomatized by the nontrivial distributive laws. 7 / 21

DRAFT In residuated lattices... A residuated binar is semilinear if it is a subdirect product of chains. All six nontrivial distributive laws hold in semilinear residuated binars. 8 / 21

DRAFT In residuated lattices... A residuated binar is semilinear if it is a subdirect product of chains. All six nontrivial distributive laws hold in semilinear residuated binars. Proposition (Blount and Tsinakis, 2003): Let A = ( A , ∧ , ∨ , · , \ , /, e ) be a residuated lattice satisfying ( x ∨ y ) ∧ e = ( x ∧ e ) ∨ ( y ∧ e ) . Then e ≤ x / y ∨ y / x iff ( / ∧ ) iff ( ∨ / ). e ≤ y \ x ∨ x \ y iff ( ∧\ ) iff ( \∨ ) 8 / 21

DRAFT The general picture In residuated binars, the above fails. 9 / 21

DRAFT The general picture In residuated binars, the above fails. Let A be the residuated binar with lattice reduct {⊥ , a , b , ⊤} , where ⊥ < a , b < ⊤ , defined in the below. 9 / 21

DRAFT The general picture In residuated binars, the above fails. Let A be the residuated binar with lattice reduct {⊥ , a , b , ⊤} , where ⊥ < a , b < ⊤ , defined in the below. · ⊥ a b ⊤ \ ⊥ a b ⊤ / ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊥ ⊤ ⊤ b b ⊥ ⊥ ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ a a a b b ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊤ b b b b b b b b ⊤ ⊥ ⊥ ⊤ ⊤ ⊤ b b b ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ 9 / 21

DRAFT The general picture In residuated binars, the above fails. Let A be the residuated binar with lattice reduct {⊥ , a , b , ⊤} , where ⊥ < a , b < ⊤ , defined in the below. · ⊥ a b ⊤ \ ⊥ a b ⊤ / ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊥ ⊤ ⊤ b b ⊥ ⊥ ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ a a a b b ⊥ ⊥ ⊤ ⊤ ⊤ ⊤ ⊤ b b b b b b b b ⊤ ⊥ ⊥ ⊤ ⊤ ⊤ b b b ⊤ ⊤ ⊤ ⊤ ⊤ ⊤ Then A | = ( ∧\ ), but A �| = ( \∨ ). 9 / 21

DRAFT The general picture (cont.) Nevertheless, there are implications among the nontrivial distributive laws... 10 / 21

DRAFT The general picture (cont.) Nevertheless, there are implications among the nontrivial distributive laws... Proposition (WF, P. Jipsen 2018): Let A be a residuated binar whose lattice reduct is distributive. Then if A satisfies both ( ∨ / ) and ( ∧\ ), A also satisfies ( \∨ ). 10 / 21

DRAFT The general picture (cont.) Nevertheless, there are implications among the nontrivial distributive laws... Proposition (WF, P. Jipsen 2018): Let A be a residuated binar whose lattice reduct is distributive. Then if A satisfies both ( ∨ / ) and ( ∧\ ), A also satisfies ( \∨ ). Proof: Note first that we can rewrite the identities ( ∧\ ) and ( \∨ ). 10 / 21

DRAFT The general picture (cont.) We have that ( ∧\ ) is equivalent to ( x ∧ y ) \ ( z ∧ w ) ≤ x \ z ∨ y \ w , and ( \∨ ) is equivalent to ( x ∨ y ) \ ( z ∨ w ) ≤ x \ z ∨ y \ w . 11 / 21

DRAFT The general picture (cont.) We have that ( ∧\ ) is equivalent to ( x ∧ y ) \ ( z ∧ w ) ≤ x \ z ∨ y \ w , and ( \∨ ) is equivalent to ( x ∨ y ) \ ( z ∨ w ) ≤ x \ z ∨ y \ w . Let u ≤ ( x ∨ y ) \ ( z ∨ w ). Then by residuation x ∨ y ≤ ( z ∨ w ) / u , and by ( ∨ / ) we have x ∨ y ≤ z / u ∨ w / u . 11 / 21

DRAFT The general picture (cont.) This gives x ∨ y = ( x ∨ y ) ∧ ( z / u ∨ w / u ). 12 / 21

DRAFT The general picture (cont.) This gives x ∨ y = ( x ∨ y ) ∧ ( z / u ∨ w / u ). Using lattice distributivity, x ∨ y = x 1 ∨ x 2 ∨ y 1 ∨ y 2 , where x 1 = x ∧ ( z / u ) , x 2 = x ∧ ( w / u ) , y 1 = y ∧ ( z / u ) , y 2 = y ∧ ( w / u ) . 12 / 21

DRAFT The general picture (cont.) This gives x ∨ y = ( x ∨ y ) ∧ ( z / u ∨ w / u ). Using lattice distributivity, x ∨ y = x 1 ∨ x 2 ∨ y 1 ∨ y 2 , where x 1 = x ∧ ( z / u ) , x 2 = x ∧ ( w / u ) , y 1 = y ∧ ( z / u ) , y 2 = y ∧ ( w / u ) . Notice: x 1 ≤ z / u = ⇒ u ≤ x 1 \ z ≤ ( x 1 ∧ y 2 ) \ z , x 2 ≤ w / u = ⇒ u ≤ x 2 \ w ≤ ( x 2 ∧ y 1 ) \ w , y 1 ≤ z / u = ⇒ u ≤ y 1 \ z ≤ ( x 2 ∧ y 1 ) \ z , y 2 ≤ y / u = ⇒ u ≤ y 2 \ w ≤ ( x 1 ∧ y 2 ) \ w . 12 / 21

DRAFT The general picture (cont.) This gives u ≤ ( x 1 ∧ y 2 ) \ ( z ∧ w ) ≤ x 1 \ z ∨ y 2 \ w u ≤ ( x 2 ∧ y 1 ) \ ( z ∧ w ) ≤ x 2 \ z ∨ y 1 \ w u ≤ x 1 \ z ≤ x 1 \ z ∨ y 1 \ w u ≤ y 2 \ w ≤ x 2 \ z ∨ y 2 \ w . 13 / 21