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Free algebras for the {↔, ¬ } fragment of IPC and KC

Free algebras for the {↔, ¬ } fragment of IPC and KC. Katarzyna Słomczyńska (Pedagogical University, Kraków) e-mail: kslomcz@ap.krakow.pl. Equivalential algebras. equivalential algebras ( E ) ⇆ equivalential fragment of intuitionistic propositional logic ( IPC ↔ ).

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Free algebras for the {↔, ¬ } fragment of IPC and KC

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  1. Free algebras for the {↔,¬}fragment of IPC and KC Katarzyna Słomczyńska (Pedagogical University, Kraków) e-mail: kslomcz@ap.krakow.pl

  2. Equivalential algebras equivalential algebras(E)⇆equivalential fragment of intuitionistic propositional logic (IPC↔ )

  3. Equivalential algebras with 0 • E0 - equivalential algebras with (constant) 0⇆{↔,¬} fragment of intuitionistic propositional logic (IPC↔,¬),¬x := x 0, 0 := ¬1 • E⇆{↔,¬} fragment of weak excluded middle logic (KC↔,¬) KC:= IPC + (¬p v ¬(¬p)= 1) • E-linear equivalential algebras with 0 ⇆{↔,¬} fragment of Gödel-Dummett logic

  4. How to construct effectively free finitely generated algebras inE0? Such constructions are known both for: • intuitionistic propositional logic (free Heyting algebras - Urquhart’73, Bellissima’86, Grigolia’87, Ghilardi’92, Butz’98, Darnière & Junker’06,Bezhanishvili’07,O’Connor’07) as well as for some of its fragments: • implication (→) (free Hilbert, or positive implication, algebras - Urquhart’74, de Bruijn’75) • implication-conjunction (→,∧) (free Brouwerian, or implicative, semilattices - de Bruijn’75, Köhler’81) • equivalence (↔) (free equivalential algebras – Wroński’93,Słomczyńska’05,Słomczyńska’08) • and many others - de Jongh, Hendriks & Renardel de Lavalette’91, Hendriks’96.

  5. Fregean varieties and frames • Fregean variety⇆1-regular + congruence orderable. • The equivalential algebras form a paradigm of congruence permutable Fregean varieties, in the sense that every such variety has a binary term that turns every of its members into an equivalential algebra (Idziak,Słomczyńska&Wroński). • Fregean frame⇆ (Cm(A), ≤, ~, ●), where A– an algebrafrom a Fregean variety. • equivalence relation: μ ~ νif and only if I [μ,μ+] and I [ν,ν+] are projective, where μ,ν∈ Cm(A) and φ+ - unique cover of φ. • Boolean group operation: Let μ ∈ (Cm(A). Defineφ ● ψ := (φ ÷ ψ )’ ∩μ+for φ,ψ ∈ μ/~ ∪ {μ+}. Then (μ/~ ∪ {μ+},●)∈E 2. • Two extreme cases (μ,ν∈ Cm(A)): • congruence distributive varieties: μ ~ νif and only if μ = ν; • equivalential algebras: μ ~ νif and only if μ+ = ν+.

  6. Fregean varieties and frames • Representation theorem: the elements of a finite algebra A from a congruence permutable Fregean variety can be represented as certain upwards closed subsets in the frame (Cm(A), ≤, ~, ●) called hereditary subsets. • To describe a finitely generated free algebra in a locally finite congruence permutable Fregean variety it suffices to find its Fregean frame. This was done for some congruence distributive varieties, as well as for Ein (Słomczyńska’08). Here we show how it works for E0 and some of its subvarieties. • F0(n) – the n-generated free equivalential algebra with 0.

  7. Construction – the frame We start from the construction of the frame of F0(n): • the frame is a poset divided into n+1layers (levels)Ek(n)(k =1,...,n+1) whose elements are labelled by proper subsets of the set {0,1,...,n}; • if two elements of the poset are comparable, then the larger lies in an upper layer; • each layer is divided into equivalence classes consisting of elements with the same successors (covers); • the top layerE1(n)consists of only one equivalence class labelled by all proper subsets of the set{0,1,...,n}and hence it has 2n+1−1 elements; • elements of any other class are labelled by all proper subsets of a subset K of {1,...,n}; • each class supplemented by K is endowed with the natural Boolean group operation (the complement of the symmetric difference with respect to K)and K is the unit of this group; • the layers are constructed inductively.

  8. Construction – the frame(top layer) The top layer E1(n)consists of only one equivalence class labelled by all proper subsets of the set{0,1,...,n} and hence it has 2n+1−1 elements. E1(2)

  9. Construction – the frame(inductive step) • Let k = 1,...,n. Assume that we have alreadydefinedthe posetE1(n)∪⋯∪Ek(n). • We consider all subsets S of the set E1(n)∪⋯∪Ek(n) suchthat: • S is an upwards closed set; • the intersection of S with each equivalence class (supplementedby the unit) is a Boolean subgroup of this class (also supplementedby the unit); • S intersects non-empty the lowest layer Ek(n) defined so far; • the labellings of all elements of S intersect non-empty and the intersection does not contain 0. • To each such set S we associate exactly one equivalenceclass in the layer Ek+1(n)labelled by all proper subsets of the intersection of the labellings ofall elements of S. The elements of this class are smaller than allelements of the set S. E1(2)

  10. Construction – the frame(inductive step) • Let k = 1,...,n. Assume that we have already defined the poset E1(n) ∪⋯∪Ek(n). • We consider all subsets S of the set E1(n) ∪⋯∪Ek(n) such that: • S is an upwards closed set; • the intersection of S with each equivalence class (supplemented by the unit) is a Boolean subgroup of this class (also supplemented by the unit); • S intersects non-empty the lowest layer Ek(n) defined so far; • the labellings of all elements of S intersect non-emptyand the intersection does not contain 0. • To each such set S we associate exactly one equivalence class in the layer Ek+1(n) labelled by all proper subsets of the intersection of the labellings of all elements of S. The elements of this class are smaller than all elements of the set S. E1(2)∪ E2(2)

  11. Construction – the frame(inductive step) • Let k = 1,...,n. Assume that we have already defined the poset E1(n) ∪⋯∪Ek(n). • We consider all subsets S of the set E1(n) ∪⋯∪Ek(n) such that: • S is an upwards closed set; • the intersection of S with each equivalence class (supplemented by the unit) is a Boolean subgroup of this class (also supplemented by the unit); • S intersects non-empty the lowest layer Ek(n) defined so far; • the labellings of all elements of S intersect non-empty and the intersection does not contain 0. • To each such set S we associate exactly one equivalence class in the layer Ek+1(n) labelled by all proper subsets of the intersection of the labellings of all elements of S. The elements of this class are smaller than all elements of the set S. E1(2)∪ E2(2) ∪ E3(2)

  12. Construction – the free algebra • Having the frame defined, we describe the universe of the freen-generated equivalential algebra as the collection of all hereditary sets, i.e., the subsets Z of the frame such that: 1.Z is an upwards closed set; 2. if all the elements larger than the elements of a given equivalenceclass are contained in Z, then the intersection of Z (supplementedby the unit) with this class is either a maximal Boolean subgroup of this class (alsosupplemented by the unit) or is equal to this class. • The family of hereditary sets is endowed with the natural equivalenceoperation of the dual pseudo-difference (i.e., for hereditary sets S and Twe define S↔T:= ((S ÷ T)↓)’. • The free equivalential algebra F 0(n) = (hereditary sets, ↔, 0), where 0 is the hereditary set consisting of all elements labelled by subsets containing 0.

  13. The sets Zk(k = 0,1,...,n) consisted of all elements having k in their labelling are 0 and free generators of the free algebraF0(n). For n = 0we haveE1(1) = 1,and 2hereditary sets. For n = 1 we have E1(1) = 3, E2(1) = 1, and 6hereditary sets. For n = 2 we have E1(2) = 7, E2(2) = 7, E3(2) = 4, and538hereditary sets. Construction – the freegenerators

  14. How to construct effectively free finitely generated algebras for subvarieties of E0? • E- frame: every equivalence class in the layer Ek+1(n) has exactlyone larger element in the top layer E1(n) • E- frame: every equivalence class in the layer Ek+1(n) has exactlyonelarger element in the layers E1(n), …, Ek(n)

  15. Free spectra for E andE

  16. Frame forE(n=3) Frame: 70 elements Free algebra: 28 170 470 730 elements

  17. Frame forE(n=3) Frame: 58 elements Free algebra: 40 706 210elements

  18. Free spectra(and the cardinalities of the frames)

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