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The setting Definition: A residuated binar is an expansion of a lattice A by binary- perations ·, \, / that satisfy the law of residuation,
DRAFT Distributivity in residuated structures Wesley Fussner - - PowerPoint PPT Presentation
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
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The setting Definition: A residuated binar is an expansion of a lattice A by binary- perations ·, \, / that satisfy the law of residuation, i.e., for all
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The setting Definition: A residuated binar is an expansion of a lattice A by binary- perations ·, \, / that satisfy the law of residuation, i.e., for all
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Residuation and distributivity One may think of the law of residuation as a “finite approximation” of an infinite distributive property. 3 / 21 SLIDE 6
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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 SLIDE 7
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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 }.
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Residuation and distributivity (cont.) The division operations \ and / also satisfy strong distributive properties. 4 / 21 SLIDE 9
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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 SLIDE 10
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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 ) =
- y∈Y
- Y )/z =
- y∈Y
- X)\z =
- x∈X
- X) =
- x∈X
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Residuation and distributivity (cont.) There is a partial converse to the above. 5 / 21 SLIDE 12
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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 SLIDE 13
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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 SLIDE 14
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Residuation and distributivity (cont.) In more finitary terms, we have: 6 / 21 SLIDE 15
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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 SLIDE 16
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Nontrivial distributive laws Consider the following nontrivial distributive laws: 7 / 21 SLIDE 17
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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 SLIDE 18
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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 SLIDE 19
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In residuated lattices... A residuated binar is semilinear if it is a subdirect product of- chains. All six nontrivial distributive laws hold in semilinear
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In residuated lattices... A residuated binar is semilinear if it is a subdirect product of- chains. All six nontrivial distributive laws hold in semilinear
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The general picture In residuated binars, the above fails. 9 / 21 SLIDE 22
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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 SLIDE 23
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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 ⊤ ⊥ ⊤ ⊤ ⊤ ⊤ a ⊤ ⊤ ⊤ ⊤ b b b b ⊤ ⊤ b b b ⊤ / ⊥ a b ⊤ ⊥ ⊤ ⊤ b b a ⊤ ⊤ b b b ⊤ ⊤ b b ⊤ ⊤ ⊤ ⊤ ⊤ 9 / 21 SLIDE 24
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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 ⊤ ⊥ ⊤ ⊤ ⊤ ⊤ a ⊤ ⊤ ⊤ ⊤ b b b b ⊤ ⊤ b b b ⊤ / ⊥ a b ⊤ ⊥ ⊤ ⊤ b b a ⊤ ⊤ b b b ⊤ ⊤ b b ⊤ ⊤ ⊤ ⊤ ⊤ Then A | = (∧\), but A | = (\∨). 9 / 21 SLIDE 25
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The general picture (cont.) Nevertheless, there are implications among the nontrivial distributive laws... 10 / 21 SLIDE 26
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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 SLIDE 27
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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 SLIDE 28
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The general picture (cont.) We have that (∧\) is equivalent to (x ∧ y)\(z ∧ w) ≤ x\z ∨ y\w, and (\∨) is equivalent to (x ∨ y)\(z ∨ w) ≤ x\z ∨ y\w. 11 / 21 SLIDE 29
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The general picture (cont.) We have that (∧\) 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 SLIDE 30
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The general picture (cont.) This gives x ∨ y = (x ∨ y) ∧ (z/u ∨ w/u). 12 / 21 SLIDE 31
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The general picture (cont.) This gives x ∨ y = (x ∨ y) ∧ (z/u ∨ w/u). Using lattice distributivity, x ∨ y = x1 ∨ x2 ∨ y1 ∨ y2, where x1 = x ∧ (z/u), x2 = x ∧ (w/u), y1 = y ∧ (z/u), y2 = y ∧ (w/u). 12 / 21 SLIDE 32
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The general picture (cont.) This gives x ∨ y = (x ∨ y) ∧ (z/u ∨ w/u). Using lattice distributivity, x ∨ y = x1 ∨ x2 ∨ y1 ∨ y2, where x1 = x ∧ (z/u), x2 = x ∧ (w/u), y1 = y ∧ (z/u), y2 = y ∧ (w/u). Notice: x1 ≤ z/u = ⇒ u ≤ x1\z ≤ (x1 ∧ y2)\z, x2 ≤ w/u = ⇒ u ≤ x2\w ≤ (x2 ∧ y1)\w, y1 ≤ z/u = ⇒ u ≤ y1\z ≤ (x2 ∧ y1)\z, y2 ≤ y/u = ⇒ u ≤ y2\w ≤ (x1 ∧ y2)\w. 12 / 21 SLIDE 33
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The general picture (cont.) This gives u ≤ (x1 ∧ y2)\(z ∧ w) ≤ x1\z ∨ y2\w u ≤ (x2 ∧ y1)\(z ∧ w) ≤ x2\z ∨ y1\w u ≤ x1\z ≤ x1\z ∨ y1\w u ≤ y2\w ≤ x2\z ∨ y2\w. 13 / 21 SLIDE 34
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The general picture (cont.) This gives u ≤ (x1 ∧ y2)\(z ∧ w) ≤ x1\z ∨ y2\w u ≤ (x2 ∧ y1)\(z ∧ w) ≤ x2\z ∨ y1\w u ≤ x1\z ≤ x1\z ∨ y1\w u ≤ y2\w ≤ x2\z ∨ y2\w. Hence, u ≤ (x1\z ∨ y2\w) ∧ (x2\z ∨ y1\w) ∧ (x1\z ∨ y1\w) ∧ (x2\z ∨ y2\w) = ((x2\z ∧ x1\z) ∨ y1\w) ∧ ((x1\z ∨ x2\z) ∨ y2\w) = (x1 ∨ x2)\z ∨ (y1\w ∧ y2\w) = x\z ∨ y\w. 13 / 21 SLIDE 35
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The general picture (cont.) By similar methods, we can obtain the following. Theorem (WF, P. Jipsen 2018): Let A be a residuated binar whose lattice reduct is distributive. If A satisfies both (∨/) and (∧\), then A also satisfies (\∨). If A satisfies both (\∨) and (/∧), then A also satisfies (∨/). If A satisfies both (·∧) and (∨/), then A also satisfies (/∧). If A satisfies both (∧·) and (\∨), then A also satisfies (∧\). If A satisfies both (∧\) and (·∧), then A also satisfies (∧·). If A satisfies both (/∧) and (∧·), then A also satisfies (·∧). 14 / 21 SLIDE 36
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The general picture (cont.) In general, there are no other implications among the nontrivial distributive laws. 15 / 21 SLIDE 37
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The general picture (cont.) In general, there are no other implications among the nontrivial distributive laws. One can construct countermodels with lattice reduct {⊥, a, b, ⊤}, ⊥ < a, b < ⊤, to demonstrate this. 15 / 21 SLIDE 38
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The general picture (cont.) In general, there are no other implications among the nontrivial distributive laws. One can construct countermodels with lattice reduct {⊥, a, b, ⊤}, ⊥ < a, b < ⊤, to demonstrate this. · ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ a ⊥ ⊥ ⊥ ⊥ b ⊥ ⊥ ⊤ ⊤ ⊤ ⊥ ⊥ ⊤ ⊤ · ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ a ⊥ ⊥ ⊥ ⊥ b ⊥ a b ⊤ ⊤ ⊥ a b ⊤ · ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ a ⊥ ⊥ a a b ⊥ ⊥ b b ⊤ ⊥ ⊥ ⊤ ⊤ · ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ a ⊥ a ⊥ a b ⊥ a ⊥ a ⊤ ⊥ a ⊥ a · ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ a ⊥ a a a b ⊥ ⊥ ⊥ ⊥ ⊤ ⊥ a a a · ⊥ a b ⊤ ⊥ ⊥ ⊥ ⊥ ⊥ a ⊥ ⊥ b b b ⊥ b ⊥ b ⊤ ⊥ b b b 15 / 21 SLIDE 39
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The role of distributivity Some remarks: 16 / 21 SLIDE 40
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The role of distributivity Some remarks: The algebraic proof above is elementary, but hard to discover. 16 / 21 SLIDE 41
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The role of distributivity Some remarks: The algebraic proof above is elementary, but hard to discover. We originally obtained the proofs of the implications between nontrivial distributive laws by duality-theoretic methods (and- ur interest in these equations stems from duality theory).
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The role of distributivity Some remarks: The algebraic proof above is elementary, but hard to discover. We originally obtained the proofs of the implications between nontrivial distributive laws by duality-theoretic methods (and- ur interest in these equations stems from duality theory).
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The role of distributivity Some remarks: The algebraic proof above is elementary, but hard to discover. We originally obtained the proofs of the implications between nontrivial distributive laws by duality-theoretic methods (and- ur interest in these equations stems from duality theory).
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The role of distributivity Some remarks: The algebraic proof above is elementary, but hard to discover. We originally obtained the proofs of the implications between nontrivial distributive laws by duality-theoretic methods (and- ur interest in these equations stems from duality theory).
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Some more results The countermodels above are all based on Boolean lattices. 17 / 21 SLIDE 46
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Some more results The countermodels above are all based on Boolean lattices. Do complements play a role? 17 / 21 SLIDE 47
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Some more results The countermodels above are all based on Boolean lattices. Do complements play a role? Proposition (WF, P. Jipsen 2018): Let A be a residuated binar with neutral element e. If e has a complement e′ and A satisfies any one of the distributive laws (·∧), (∧·), (∧\), (/∧), then A is integral (i.e., A satisfies x ≤ e). 17 / 21 SLIDE 48
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Some more results Proof: We prove the claim for (·∧). 18 / 21 SLIDE 49
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Some more results Proof: We prove the claim for (·∧). Note that ⊥ = ⊤ · ⊥ = (e ∨ e′)(e ∧ e′) = [(e ∨ e′)e] ∧ [(e ∨ e′)e′] = ⊤ ∧ (ee′ ∨ (e′)2) = e′ ∨ (e′)2 18 / 21 SLIDE 50
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Some more results Proof: We prove the claim for (·∧). Note that ⊥ = ⊤ · ⊥ = (e ∨ e′)(e ∧ e′) = [(e ∨ e′)e] ∧ [(e ∨ e′)e′] = ⊤ ∧ (ee′ ∨ (e′)2) = e′ ∨ (e′)2 Thus e′ = (e′)2 = ⊥, and hence that ⊥ is a complement of e. It follows that e = e ∨ ⊥ = ⊤. 18 / 21 SLIDE 51
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Some more results (cont.) Proposition (???): Let A be a complemented residuated binar with neutral element e. If A is integral, then ∧ and · coincide. 19 / 21 SLIDE 52
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Some more results (cont.) Proposition (???): Let A be a complemented residuated binar with neutral element e. If A is integral, then ∧ and · coincide. Corollary: Let A be a complemented residuated binar with neutral element e. If A satisfies any one of the distributive laws (·∧), (∧·), (∧\), (/∧), then A is a Boolean algebra. 19 / 21 SLIDE 53
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Concluding remarks Work is still on-going, and many questions remain: 20 / 21 SLIDE 54
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Concluding remarks Work is still on-going, and many questions remain: The non-distributive case is still open. 20 / 21 SLIDE 55
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Concluding remarks Work is still on-going, and many questions remain: The non-distributive case is still open. The role of semilinearity in the absence of associativity and a multiplicative neutral element is also unknown. 20 / 21 SLIDE 56
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Concluding remarks Work is still on-going, and many questions remain: The non-distributive case is still open. The role of semilinearity in the absence of associativity and a multiplicative neutral element is also unknown. What is the join of two varieties axiomatized by a collection of nontrivial distributive laws? 20 / 21 SLIDE 57