FirstOrder Logics and Truth Degrees


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1 FirstOrder Logics and Truth Degrees George Metcalfe Mathematics Institute University of Bern LATD 2014, Vienna Summer of Logic, July 2014 George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
2 The Inebriation Principle There is someone at the party such that if he or she is drunk, then everyone is drunk. ( x)(d(x) ( y)d(y)) Either everyone is drunk and we choose anyone, or someone is not drunk and classically if this person is drunk, then everyone is drunk. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
3 The Inebriation Principle There is someone at the party such that if he or she is drunk, then everyone is drunk. ( x)(d(x) ( y)d(y)) Either everyone is drunk and we choose anyone, or someone is not drunk and classically if this person is drunk, then everyone is drunk. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
4 The Inebriation Principle There is someone at the party such that if he or she is drunk, then everyone is drunk. ( x)(d(x) ( y)d(y)) Either everyone is drunk and we choose anyone, or someone is not drunk and classically if this person is drunk, then everyone is drunk. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
5 Classical Inebriation By prenexation, ( x)(d(x) ( y)d(y)) is equivalent to ( x)( y)(d(x) D(y)), which, by Skolemization, is valid if and only if ( x)(d(x) D(f (x))) is valid, which, using Herbrand s theorem, it is, because (D(c) D(f (c))) (D(f (c)) D(f (f (c)))) is propositionally valid in classical logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
6 Classical Inebriation By prenexation, ( x)(d(x) ( y)d(y)) is equivalent to ( x)( y)(d(x) D(y)), which, by Skolemization, is valid if and only if ( x)(d(x) D(f (x))) is valid, which, using Herbrand s theorem, it is, because (D(c) D(f (c))) (D(f (c)) D(f (f (c)))) is propositionally valid in classical logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
7 Classical Inebriation By prenexation, ( x)(d(x) ( y)d(y)) is equivalent to ( x)( y)(d(x) D(y)), which, by Skolemization, is valid if and only if ( x)(d(x) D(f (x))) is valid, which, using Herbrand s theorem, it is, because (D(c) D(f (c))) (D(f (c)) D(f (f (c)))) is propositionally valid in classical logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
8 ManyValued Inebriation How does our inebriation principle fare with respect to truth values in [0, 1]... truth as 1 and falsity as 0... as minimum and as maximum... as inf and as sup? Roughly, the principle holds for finitely many truth values, but for infinitely many values, it depends on how we interpret implication. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
9 ManyValued Inebriation How does our inebriation principle fare with respect to truth values in [0, 1]... truth as 1 and falsity as 0... as minimum and as maximum... as inf and as sup? Roughly, the principle holds for finitely many truth values, but for infinitely many values, it depends on how we interpret implication. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
10 ManyValued Inebriation How does our inebriation principle fare with respect to truth values in [0, 1]... truth as 1 and falsity as 0... as minimum and as maximum... as inf and as sup? Roughly, the principle holds for finitely many truth values, but for infinitely many values, it depends on how we interpret implication. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
11 ManyValued Inebriation How does our inebriation principle fare with respect to truth values in [0, 1]... truth as 1 and falsity as 0... as minimum and as maximum... as inf and as sup? Roughly, the principle holds for finitely many truth values, but for infinitely many values, it depends on how we interpret implication. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
12 ManyValued Inebriation How does our inebriation principle fare with respect to truth values in [0, 1]... truth as 1 and falsity as 0... as minimum and as maximum... as inf and as sup? Roughly, the principle holds for finitely many truth values, but for infinitely many values, it depends on how we interpret implication. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
13 ManyValued Inebriation How does our inebriation principle fare with respect to truth values in [0, 1]... truth as 1 and falsity as 0... as minimum and as maximum... as inf and as sup? Roughly, the principle holds for finitely many truth values, but for infinitely many values, it depends on how we interpret implication. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
14 Two Logics Gödel vs Lukasiewicz George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
15 Ordered Inebriation Given truth values [0, 1] and the Gödel implication { 1 if a b a b = b otherwise, the inebriation principle ( x)(d(x) ( y)d(y)) fails. Suppose that there are people p 1, p 2,... and that each person p n is drunk to degree 1 n. Then everyone is drunk is completely false, as is p n is drunk implies everyone is drunk. So the inebriation principle has degree 0. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
16 Ordered Inebriation Given truth values [0, 1] and the Gödel implication { 1 if a b a b = b otherwise, the inebriation principle ( x)(d(x) ( y)d(y)) fails. Suppose that there are people p 1, p 2,... and that each person p n is drunk to degree 1 n. Then everyone is drunk is completely false, as is p n is drunk implies everyone is drunk. So the inebriation principle has degree 0. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
17 Ordered Inebriation Given truth values [0, 1] and the Gödel implication { 1 if a b a b = b otherwise, the inebriation principle ( x)(d(x) ( y)d(y)) fails. Suppose that there are people p 1, p 2,... and that each person p n is drunk to degree 1 n. Then everyone is drunk is completely false, as is p n is drunk implies everyone is drunk. So the inebriation principle has degree 0. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
18 Ordered Inebriation Given truth values [0, 1] and the Gödel implication { 1 if a b a b = b otherwise, the inebriation principle ( x)(d(x) ( y)d(y)) fails. Suppose that there are people p 1, p 2,... and that each person p n is drunk to degree 1 n. Then everyone is drunk is completely false, as is p n is drunk implies everyone is drunk. So the inebriation principle has degree 0. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
19 Ordered Inebriation Given truth values [0, 1] and the Gödel implication { 1 if a b a b = b otherwise, the inebriation principle ( x)(d(x) ( y)d(y)) fails. Suppose that there are people p 1, p 2,... and that each person p n is drunk to degree 1 n. Then everyone is drunk is completely false, as is p n is drunk implies everyone is drunk. So the inebriation principle has degree 0. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
20 Ordered Inebriation Given truth values [0, 1] and the Gödel implication { 1 if a b a b = b otherwise, the inebriation principle ( x)(d(x) ( y)d(y)) fails. Suppose that there are people p 1, p 2,... and that each person p n is drunk to degree 1 n. Then everyone is drunk is completely false, as is p n is drunk implies everyone is drunk. So the inebriation principle has degree 0. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
21 Continuous Inebriation Given truth values [0, 1] and the Lukasiewicz implication a b = min(1, 1 a + b), the inebriation principle ( x)(d(x) ( y)d(y)) holds. Suppose that everybody is drunk to at least degree k and that k is the greatest truth value with this property. Then for each ɛ > 0, there is a person p ɛ drunk to at most degree k + ɛ and p ɛ is drunk implies everyone is drunk has degree at least 1 ɛ. So the inebriation principle has value 1. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
22 Continuous Inebriation Given truth values [0, 1] and the Lukasiewicz implication a b = min(1, 1 a + b), the inebriation principle ( x)(d(x) ( y)d(y)) holds. Suppose that everybody is drunk to at least degree k and that k is the greatest truth value with this property. Then for each ɛ > 0, there is a person p ɛ drunk to at most degree k + ɛ and p ɛ is drunk implies everyone is drunk has degree at least 1 ɛ. So the inebriation principle has value 1. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
23 Continuous Inebriation Given truth values [0, 1] and the Lukasiewicz implication a b = min(1, 1 a + b), the inebriation principle ( x)(d(x) ( y)d(y)) holds. Suppose that everybody is drunk to at least degree k and that k is the greatest truth value with this property. Then for each ɛ > 0, there is a person p ɛ drunk to at most degree k + ɛ and p ɛ is drunk implies everyone is drunk has degree at least 1 ɛ. So the inebriation principle has value 1. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
24 Continuous Inebriation Given truth values [0, 1] and the Lukasiewicz implication a b = min(1, 1 a + b), the inebriation principle ( x)(d(x) ( y)d(y)) holds. Suppose that everybody is drunk to at least degree k and that k is the greatest truth value with this property. Then for each ɛ > 0, there is a person p ɛ drunk to at most degree k + ɛ and p ɛ is drunk implies everyone is drunk has degree at least 1 ɛ. So the inebriation principle has value 1. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
25 Continuous Inebriation Given truth values [0, 1] and the Lukasiewicz implication a b = min(1, 1 a + b), the inebriation principle ( x)(d(x) ( y)d(y)) holds. Suppose that everybody is drunk to at least degree k and that k is the greatest truth value with this property. Then for each ɛ > 0, there is a person p ɛ drunk to at most degree k + ɛ and p ɛ is drunk implies everyone is drunk has degree at least 1 ɛ. So the inebriation principle has value 1. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
26 Continuous Inebriation Given truth values [0, 1] and the Lukasiewicz implication a b = min(1, 1 a + b), the inebriation principle ( x)(d(x) ( y)d(y)) holds. Suppose that everybody is drunk to at least degree k and that k is the greatest truth value with this property. Then for each ɛ > 0, there is a person p ɛ drunk to at most degree k + ɛ and p ɛ is drunk implies everyone is drunk has degree at least 1 ɛ. So the inebriation principle has value 1. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
27 This Talk Logical consequence in firstorder classical logic, Γ = ϕ, reduces to the unsatisfiability of Γ { ϕ} via doublenegation and the deduction theorem a set of prenex formulas via quantifier shifts a set of universal formulas via Skolemization a set of propositional formulas via Herbrand s theorem. Main Question Which of these steps can be applied in the manyvalued setting? George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
28 This Talk Logical consequence in firstorder classical logic, Γ = ϕ, reduces to the unsatisfiability of Γ { ϕ} via doublenegation and the deduction theorem a set of prenex formulas via quantifier shifts a set of universal formulas via Skolemization a set of propositional formulas via Herbrand s theorem. Main Question Which of these steps can be applied in the manyvalued setting? George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
29 This Talk Logical consequence in firstorder classical logic, Γ = ϕ, reduces to the unsatisfiability of Γ { ϕ} via doublenegation and the deduction theorem a set of prenex formulas via quantifier shifts a set of universal formulas via Skolemization a set of propositional formulas via Herbrand s theorem. Main Question Which of these steps can be applied in the manyvalued setting? George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
30 This Talk Logical consequence in firstorder classical logic, Γ = ϕ, reduces to the unsatisfiability of Γ { ϕ} via doublenegation and the deduction theorem a set of prenex formulas via quantifier shifts a set of universal formulas via Skolemization a set of propositional formulas via Herbrand s theorem. Main Question Which of these steps can be applied in the manyvalued setting? George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
31 This Talk Logical consequence in firstorder classical logic, Γ = ϕ, reduces to the unsatisfiability of Γ { ϕ} via doublenegation and the deduction theorem a set of prenex formulas via quantifier shifts a set of universal formulas via Skolemization a set of propositional formulas via Herbrand s theorem. Main Question Which of these steps can be applied in the manyvalued setting? George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
32 This Talk Logical consequence in firstorder classical logic, Γ = ϕ, reduces to the unsatisfiability of Γ { ϕ} via doublenegation and the deduction theorem a set of prenex formulas via quantifier shifts a set of universal formulas via Skolemization a set of propositional formulas via Herbrand s theorem. Main Question Which of these steps can be applied in the manyvalued setting? George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
33 Disclaimer This talk is based partly on P. Cintula and G. Metcalfe. Herbrand Theorems for Substructural Logics. Proceedings of LPAR 2013, LNCS 8312, Springer, For further details and references see George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
34 Beyond Two Values We consider a propositional language L and (classes of) Lalgebras A = A,,, { i } i I,, where A,,,, is a complete chain (e.g., {0, 1}, [0, 1], N { }). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
35 OrderBased Algebras For Gödel implication on A, only the order of values matters... { if a b a b = b otherwise and similarly for operations such as { if a = a = and a b = otherwise { b if b a otherwise. More formally, such operations are definable in A by a quantifierfree formula in the firstorder language with only,, and constants of L. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
36 OrderBased Algebras For Gödel implication on A, only the order of values matters... { if a b a b = b otherwise and similarly for operations such as { if a = a = and a b = otherwise { b if b a otherwise. More formally, such operations are definable in A by a quantifierfree formula in the firstorder language with only,, and constants of L. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
37 OrderBased Algebras For Gödel implication on A, only the order of values matters... { if a b a b = b otherwise and similarly for operations such as { if a = a = and a b = otherwise { b if b a otherwise. More formally, such operations are definable in A by a quantifierfree formula in the firstorder language with only,, and constants of L. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
38 The Standard Lukasiewicz Algebra Lukasiewicz implication on [0, 1] is the continuous function a b = min(1, 1 a + b). The functions min and max are definable using and 0, as are a = 1 a and a b = min(1, a + b). Indeed, interpretations of formulas relate 11 to piecewise linear continuous functions on [0, 1] with integer coefficients (McNaughton 1951). Other logics of continuous functions include rational Pavelka logic, continuous logic, and abelian logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
39 The Standard Lukasiewicz Algebra Lukasiewicz implication on [0, 1] is the continuous function a b = min(1, 1 a + b). The functions min and max are definable using and 0, as are a = 1 a and a b = min(1, a + b). Indeed, interpretations of formulas relate 11 to piecewise linear continuous functions on [0, 1] with integer coefficients (McNaughton 1951). Other logics of continuous functions include rational Pavelka logic, continuous logic, and abelian logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
40 The Standard Lukasiewicz Algebra Lukasiewicz implication on [0, 1] is the continuous function a b = min(1, 1 a + b). The functions min and max are definable using and 0, as are a = 1 a and a b = min(1, a + b). Indeed, interpretations of formulas relate 11 to piecewise linear continuous functions on [0, 1] with integer coefficients (McNaughton 1951). Other logics of continuous functions include rational Pavelka logic, continuous logic, and abelian logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
41 The Standard Lukasiewicz Algebra Lukasiewicz implication on [0, 1] is the continuous function a b = min(1, 1 a + b). The functions min and max are definable using and 0, as are a = 1 a and a b = min(1, a + b). Indeed, interpretations of formulas relate 11 to piecewise linear continuous functions on [0, 1] with integer coefficients (McNaughton 1951). Other logics of continuous functions include rational Pavelka logic, continuous logic, and abelian logic. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
42 FirstOrder Languages A predicate language P is a triple P, F, ar where P and F are nonempty sets of predicate symbols and function symbols, respectively, and ar is a function assigning to P F an arity ar( ) N. Pterms s, t,... and Pformulas ϕ, ψ,... are defined as in classical logic using a fixed countably infinite set OV of object variables x, y,..., propositional connectives from L, and the quantifiers and. A Ptheory Γ is just a set of Pformulas. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
43 FirstOrder Languages A predicate language P is a triple P, F, ar where P and F are nonempty sets of predicate symbols and function symbols, respectively, and ar is a function assigning to P F an arity ar( ) N. Pterms s, t,... and Pformulas ϕ, ψ,... are defined as in classical logic using a fixed countably infinite set OV of object variables x, y,..., propositional connectives from L, and the quantifiers and. A Ptheory Γ is just a set of Pformulas. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
44 FirstOrder Languages A predicate language P is a triple P, F, ar where P and F are nonempty sets of predicate symbols and function symbols, respectively, and ar is a function assigning to P F an arity ar( ) N. Pterms s, t,... and Pformulas ϕ, ψ,... are defined as in classical logic using a fixed countably infinite set OV of object variables x, y,..., propositional connectives from L, and the quantifiers and. A Ptheory Γ is just a set of Pformulas. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
45 FirstOrder Structures A Pstructure M = A, M consists of an Lalgebra A based on a complete chain, and M = M, {P M } P P, {f M } f F where M is a nonempty set, P M : M n A is a function for each nary P P, and f M : M n M is a function for each nary f F. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
46 Evaluations Given an Mevaluation v mapping object variables to M, x M v = v(x) (x OV ) f (t 1,..., t n ) M v = f M ( t 1 M v,..., t n M v P(t 1,..., t n ) M v = P M ( t 1 M v,..., t n M v (ϕ 1,..., ϕ n ) M v = A ( ϕ 1 M v,..., ϕ n M v ) (f F) ) (P P) ) ( L), and letting v[x a](x) = a and v[x a](y) = v(y) for y x, ( x)ϕ M v = A a M ϕ M v[x a] ( x)ϕ M v = A a M ϕ M v[x a]. Notation. Given v( x) = a, we often write ϕ( a) M for ϕ( x) M v. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
47 Evaluations Given an Mevaluation v mapping object variables to M, x M v = v(x) (x OV ) f (t 1,..., t n ) M v = f M ( t 1 M v,..., t n M v P(t 1,..., t n ) M v = P M ( t 1 M v,..., t n M v (ϕ 1,..., ϕ n ) M v = A ( ϕ 1 M v,..., ϕ n M v ) (f F) ) (P P) ) ( L), and letting v[x a](x) = a and v[x a](y) = v(y) for y x, ( x)ϕ M v = A a M ϕ M v[x a] ( x)ϕ M v = A a M ϕ M v[x a]. Notation. Given v( x) = a, we often write ϕ( a) M for ϕ( x) M v. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
48 Evaluations Given an Mevaluation v mapping object variables to M, x M v = v(x) (x OV ) f (t 1,..., t n ) M v = f M ( t 1 M v,..., t n M v P(t 1,..., t n ) M v = P M ( t 1 M v,..., t n M v (ϕ 1,..., ϕ n ) M v = A ( ϕ 1 M v,..., ϕ n M v ) (f F) ) (P P) ) ( L), and letting v[x a](x) = a and v[x a](y) = v(y) for y x, ( x)ϕ M v = A a M ϕ M v[x a] ( x)ϕ M v = A a M ϕ M v[x a]. Notation. Given v( x) = a, we often write ϕ( a) M for ϕ( x) M v. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
49 Evaluations Given an Mevaluation v mapping object variables to M, x M v = v(x) (x OV ) f (t 1,..., t n ) M v = f M ( t 1 M v,..., t n M v P(t 1,..., t n ) M v = P M ( t 1 M v,..., t n M v (ϕ 1,..., ϕ n ) M v = A ( ϕ 1 M v,..., ϕ n M v ) (f F) ) (P P) ) ( L), and letting v[x a](x) = a and v[x a](y) = v(y) for y x, ( x)ϕ M v = A a M ϕ M v[x a] ( x)ϕ M v = A a M ϕ M v[x a]. Notation. Given v( x) = a, we often write ϕ( a) M for ϕ( x) M v. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
50 Evaluations Given an Mevaluation v mapping object variables to M, x M v = v(x) (x OV ) f (t 1,..., t n ) M v = f M ( t 1 M v,..., t n M v P(t 1,..., t n ) M v = P M ( t 1 M v,..., t n M v (ϕ 1,..., ϕ n ) M v = A ( ϕ 1 M v,..., ϕ n M v ) (f F) ) (P P) ) ( L), and letting v[x a](x) = a and v[x a](y) = v(y) for y x, ( x)ϕ M v = A a M ϕ M v[x a] ( x)ϕ M v = A a M ϕ M v[x a]. Notation. Given v( x) = a, we often write ϕ( a) M for ϕ( x) M v. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
51 Models Let K be a class of Lalgebras based on complete chains and let A K. Then a Pstructure M = A, M is a Kmodel of a Ptheory Γ, written M = Γ, if ϕ M v = A for each ϕ Γ and Mevaluation v. Note. A Pstructure M = A, M is called witnessed if for each Pformula ϕ(x, y) and a M, there exist b, c M such that ( x)ϕ(x, a) M = ϕ(b, a) M and ( x)ϕ(x, a) M = ϕ(c, a) M. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
52 Models Let K be a class of Lalgebras based on complete chains and let A K. Then a Pstructure M = A, M is a Kmodel of a Ptheory Γ, written M = Γ, if ϕ M v = A for each ϕ Γ and Mevaluation v. Note. A Pstructure M = A, M is called witnessed if for each Pformula ϕ(x, y) and a M, there exist b, c M such that ( x)ϕ(x, a) M = ϕ(b, a) M and ( x)ϕ(x, a) M = ϕ(c, a) M. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
53 Models Let K be a class of Lalgebras based on complete chains and let A K. Then a Pstructure M = A, M is a Kmodel of a Ptheory Γ, written M = Γ, if ϕ M v = A for each ϕ Γ and Mevaluation v. Note. A Pstructure M = A, M is called witnessed if for each Pformula ϕ(x, y) and a M, there exist b, c M such that ( x)ϕ(x, a) M = ϕ(b, a) M and ( x)ϕ(x, a) M = ϕ(c, a) M. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
54 Consequence If Γ {ϕ} is a Ptheory and for each A K and Pstructure M = A, M, M = Γ = M = ϕ, then we say that ϕ is a consequence of Γ in K, written Γ = K ϕ. In particular, if = K ϕ, then ϕ is said to be valid in K. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
55 Consequence If Γ {ϕ} is a Ptheory and for each A K and Pstructure M = A, M, M = Γ = M = ϕ, then we say that ϕ is a consequence of Γ in K, written Γ = K ϕ. In particular, if = K ϕ, then ϕ is said to be valid in K. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
56 FirstOrder Gödel Logic When K consists of just the standard Gödel algebra on [0, 1], we obtain firstorder Gödel logic (and write = G ) which admits the deduction theorem Γ {ϕ} = G ψ Γ = G ϕ ψ (ϕ, ψ sentences) but not double negation, i.e., = G ϕ ϕ admits some quantifier shifts such as = G (( x)ϕ ψ) ( x)(ϕ ψ) (x not free in ψ) but not others, e.g., = G (ϕ ( x)ψ) ( x)(ϕ ψ) (x not free in ϕ) George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
57 FirstOrder Gödel Logic When K consists of just the standard Gödel algebra on [0, 1], we obtain firstorder Gödel logic (and write = G ) which admits the deduction theorem Γ {ϕ} = G ψ Γ = G ϕ ψ (ϕ, ψ sentences) but not double negation, i.e., = G ϕ ϕ admits some quantifier shifts such as = G (( x)ϕ ψ) ( x)(ϕ ψ) (x not free in ψ) but not others, e.g., = G (ϕ ( x)ψ) ( x)(ϕ ψ) (x not free in ϕ) George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
58 FirstOrder Gödel Logic When K consists of just the standard Gödel algebra on [0, 1], we obtain firstorder Gödel logic (and write = G ) which admits the deduction theorem Γ {ϕ} = G ψ Γ = G ϕ ψ (ϕ, ψ sentences) but not double negation, i.e., = G ϕ ϕ admits some quantifier shifts such as = G (( x)ϕ ψ) ( x)(ϕ ψ) (x not free in ψ) but not others, e.g., = G (ϕ ( x)ψ) ( x)(ϕ ψ) (x not free in ϕ) George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
59 FirstOrder Gödel Logic When K consists of just the standard Gödel algebra on [0, 1], we obtain firstorder Gödel logic (and write = G ) which admits the deduction theorem Γ {ϕ} = G ψ Γ = G ϕ ψ (ϕ, ψ sentences) but not double negation, i.e., = G ϕ ϕ admits some quantifier shifts such as = G (( x)ϕ ψ) ( x)(ϕ ψ) (x not free in ψ) but not others, e.g., = G (ϕ ( x)ψ) ( x)(ϕ ψ) (x not free in ϕ) George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
60 FirstOrder Gödel Logic When K consists of just the standard Gödel algebra on [0, 1], we obtain firstorder Gödel logic (and write = G ) which admits the deduction theorem Γ {ϕ} = G ψ Γ = G ϕ ψ (ϕ, ψ sentences) but not double negation, i.e., = G ϕ ϕ admits some quantifier shifts such as = G (( x)ϕ ψ) ( x)(ϕ ψ) (x not free in ψ) but not others, e.g., = G (ϕ ( x)ψ) ( x)(ϕ ψ) (x not free in ϕ) George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
61 FirstOrder Gödel Logic is (recursively) axiomatizable (Horn 1969) as the extension of firstorder intuitionistic logic with is finitary, i.e., (prl) (ϕ ψ) (ψ ϕ) (cd) ( x)(ϕ ψ) (ϕ ( x)ψ) (x not free in ϕ) Γ = G ϕ Γ = G ϕ for some finite Γ Γ but not complete with respect to witnessed models has an elegant proof theory (Avron 1991, Baaz & Zach 2000). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
62 FirstOrder Gödel Logic is (recursively) axiomatizable (Horn 1969) as the extension of firstorder intuitionistic logic with is finitary, i.e., (prl) (ϕ ψ) (ψ ϕ) (cd) ( x)(ϕ ψ) (ϕ ( x)ψ) (x not free in ϕ) Γ = G ϕ Γ = G ϕ for some finite Γ Γ but not complete with respect to witnessed models has an elegant proof theory (Avron 1991, Baaz & Zach 2000). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
63 FirstOrder Gödel Logic is (recursively) axiomatizable (Horn 1969) as the extension of firstorder intuitionistic logic with is finitary, i.e., (prl) (ϕ ψ) (ψ ϕ) (cd) ( x)(ϕ ψ) (ϕ ( x)ψ) (x not free in ϕ) Γ = G ϕ Γ = G ϕ for some finite Γ Γ but not complete with respect to witnessed models has an elegant proof theory (Avron 1991, Baaz & Zach 2000). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
64 FirstOrder Gödel Logic is (recursively) axiomatizable (Horn 1969) as the extension of firstorder intuitionistic logic with is finitary, i.e., (prl) (ϕ ψ) (ψ ϕ) (cd) ( x)(ϕ ψ) (ϕ ( x)ψ) (x not free in ϕ) Γ = G ϕ Γ = G ϕ for some finite Γ Γ but not complete with respect to witnessed models has an elegant proof theory (Avron 1991, Baaz & Zach 2000). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
65 FirstOrder Gödel Logics Remark. Restricting the standard Gödel algebra to a closed infinite subset of [0, 1] containing {0, 1}, we obtain countably infinitely many different firstorder Gödel logics (Beckmann, Goldstern, and Preining 2008). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
66 FirstOrder Lukasiewicz Logic When K consists of the standard Lukasiewicz algebra on [0, 1], we obtain firstorder Lukasiewicz logic (and write = Ł ), which admits double negation, but not the deduction theorem admits all quantifier shifts and therefore has prenexation is not finitary, but is complete with respect to witnessed models George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
67 FirstOrder Lukasiewicz Logic When K consists of the standard Lukasiewicz algebra on [0, 1], we obtain firstorder Lukasiewicz logic (and write = Ł ), which admits double negation, but not the deduction theorem admits all quantifier shifts and therefore has prenexation is not finitary, but is complete with respect to witnessed models George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
68 FirstOrder Lukasiewicz Logic When K consists of the standard Lukasiewicz algebra on [0, 1], we obtain firstorder Lukasiewicz logic (and write = Ł ), which admits double negation, but not the deduction theorem admits all quantifier shifts and therefore has prenexation is not finitary, but is complete with respect to witnessed models George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
69 FirstOrder Lukasiewicz Logic When K consists of the standard Lukasiewicz algebra on [0, 1], we obtain firstorder Lukasiewicz logic (and write = Ł ), which admits double negation, but not the deduction theorem admits all quantifier shifts and therefore has prenexation is not finitary, but is complete with respect to witnessed models George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
70 FirstOrder Lukasiewicz Logic is not (recursively) axiomatizable (Scarpellini 1962), and indeed is Π 2 complete (Ragaz 1981) has an elegant proof theory with an infinitary rule (Metcalfe, Olivetti, and Gabbay 2005, Baaz and Metcalfe 2009) provides a basis for continuous model theory (Ben Yaacov 2008). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
71 FirstOrder Lukasiewicz Logic is not (recursively) axiomatizable (Scarpellini 1962), and indeed is Π 2 complete (Ragaz 1981) has an elegant proof theory with an infinitary rule (Metcalfe, Olivetti, and Gabbay 2005, Baaz and Metcalfe 2009) provides a basis for continuous model theory (Ben Yaacov 2008). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
72 FirstOrder Lukasiewicz Logic is not (recursively) axiomatizable (Scarpellini 1962), and indeed is Π 2 complete (Ragaz 1981) has an elegant proof theory with an infinitary rule (Metcalfe, Olivetti, and Gabbay 2005, Baaz and Metcalfe 2009) provides a basis for continuous model theory (Ben Yaacov 2008). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
73 Other Possibilities... If K consists of the standard product algebra on [0, 1] with { 1 a b a b = ab and a b = otherwise, then we obtain firstorder product logic. If K consists of the complete chains of certain varieties of integral commutative residuated lattices, then we obtain a (recursively) axiomatizable and finitary firstorder logic (e.g., firstorder MTL). b a George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
74 Other Possibilities... If K consists of the standard product algebra on [0, 1] with { 1 a b a b = ab and a b = otherwise, then we obtain firstorder product logic. If K consists of the complete chains of certain varieties of integral commutative residuated lattices, then we obtain a (recursively) axiomatizable and finitary firstorder logic (e.g., firstorder MTL). b a George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
75 HalfTime Score Classical Gödel Lukasiewicz Product Double Negation YES NO YES NO Deduction Theorem YES YES NO NO Prenex Forms YES NO YES NO Witnessed Models YES NO YES NO Axiomatizable YES YES NO NO Finitary YES YES NO NO George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
76 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
77 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
78 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
79 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
80 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
81 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
82 Skolemization Right Theorem Let K be a class of Lalgebras based on complete chains. For any Ptheory Γ {ϕ(x, y)} and function symbol f ϕ P with arity y : Γ = K ( y)( x)ϕ(x, y) Γ = K ( y)ϕ(f ϕ ( y), y). Proof. ( ) Easy, because ( y)( x)ϕ(x, y) = K ( y)ϕ(f ϕ ( y), y). ( ) Suppose Γ = K ( y)ϕ(f ϕ ( y), y): i.e., ( y)( x)ϕ(x, y) M < A for some A K and model M = A, M of Γ. Then ( y)( x)ϕ(x, y) M r for some r < A. So for each m M, there is a d M satisfying ϕ(d, m) M r. Now define (using the axiom of choice) f ϕ ( m) = d with ϕ(d, m) M r, giving ( y)ϕ(f ϕ ( y), y) M r < A. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
83 Skolemization Left Theorem Let K consist of the standard Lukasiewicz or Gödel algebra. For any Ptheory Γ {ϕ(x, y), ψ} and function symbol f ϕ P with arity y : Γ {( y)( x)ϕ(x, y)} = K ψ Γ {( y)ϕ(f ϕ ( y), y)} = K ψ. Moreover, firstorder Lukasiewicz logic has prenexation, so in this case we can Skolemize all formulas on both sides of the consequence relation. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
84 Skolemization Left Theorem Let K consist of the standard Lukasiewicz or Gödel algebra. For any Ptheory Γ {ϕ(x, y), ψ} and function symbol f ϕ P with arity y : Γ {( y)( x)ϕ(x, y)} = K ψ Γ {( y)ϕ(f ϕ ( y), y)} = K ψ. Moreover, firstorder Lukasiewicz logic has prenexation, so in this case we can Skolemize all formulas on both sides of the consequence relation. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
85 Skolemization Left and Regular Completions A class K of integral commutative residuated lattices admits regular completions if for each A K, there is an embedding of A into some complete B K that preserves infinite meets and joins when they exist. Theorem Let K be the class of complete chains of a variety of integral commutative residuated lattices whose class of chains admits regular completions. For any Ptheory Γ {ϕ(x, y), ψ} and function symbol f ϕ P with arity y : Γ {( y)( x)ϕ(x, y)} = K ψ Γ {( y)ϕ(f ϕ ( y), y)} = K ψ. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
86 Skolemization Left and Regular Completions A class K of integral commutative residuated lattices admits regular completions if for each A K, there is an embedding of A into some complete B K that preserves infinite meets and joins when they exist. Theorem Let K be the class of complete chains of a variety of integral commutative residuated lattices whose class of chains admits regular completions. For any Ptheory Γ {ϕ(x, y), ψ} and function symbol f ϕ P with arity y : Γ {( y)( x)ϕ(x, y)} = K ψ Γ {( y)ϕ(f ϕ ( y), y)} = K ψ. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
87 Skolemization Left and Regular Completions A class K of integral commutative residuated lattices admits regular completions if for each A K, there is an embedding of A into some complete B K that preserves infinite meets and joins when they exist. Theorem Let K be the class of complete chains of a variety of integral commutative residuated lattices whose class of chains admits regular completions. For any Ptheory Γ {ϕ(x, y), ψ} and function symbol f ϕ P with arity y : Γ {( y)( x)ϕ(x, y)} = K ψ Γ {( y)ϕ(f ϕ ( y), y)} = K ψ. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
88 A Herbrand Theorem for FirstOrder Gödel Logic The Herbrand universe U(P) for a predicate language P with at least one constant is the set of ground (variablefree) Pterms. Theorem For any quantifierfree Pformula ϕ( x): n = G ( x)ϕ( x) = G ϕ( t i ) for some t 1,..., t n U(P). i=1 This has been proved prooftheoretically by Baaz and Zach (2000) and semantically by Baaz, Ciabattoni, and Fermüller (2001). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
89 A Herbrand Theorem for FirstOrder Gödel Logic The Herbrand universe U(P) for a predicate language P with at least one constant is the set of ground (variablefree) Pterms. Theorem For any quantifierfree Pformula ϕ( x): n = G ( x)ϕ( x) = G ϕ( t i ) for some t 1,..., t n U(P). i=1 This has been proved prooftheoretically by Baaz and Zach (2000) and semantically by Baaz, Ciabattoni, and Fermüller (2001). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
90 An Expansion Lemma Let 0 denote the quantifierfree Pformulas, and define using BNF: guniversal formulas P ::= 0 P P P P ( x)p N P gexistential formulas N ::= 0 N N N N ( x)n P N. We refer to theories containing only guniversal and gexistential formulas as guniversal and gexistential theories, respectively. Lemma For each gexistential Pformula ψ and guniversal Ptheory Γ {ϕ}: Γ {( x)ϕ( x)} = G ψ Γ {ϕ( t) t U(P)} = G ψ. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
91 An Expansion Lemma Let 0 denote the quantifierfree Pformulas, and define using BNF: guniversal formulas P ::= 0 P P P P ( x)p N P gexistential formulas N ::= 0 N N N N ( x)n P N. We refer to theories containing only guniversal and gexistential formulas as guniversal and gexistential theories, respectively. Lemma For each gexistential Pformula ψ and guniversal Ptheory Γ {ϕ}: Γ {( x)ϕ( x)} = G ψ Γ {ϕ( t) t U(P)} = G ψ. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
92 An Expansion Lemma Let 0 denote the quantifierfree Pformulas, and define using BNF: guniversal formulas P ::= 0 P P P P ( x)p N P gexistential formulas N ::= 0 N N N N ( x)n P N. We refer to theories containing only guniversal and gexistential formulas as guniversal and gexistential theories, respectively. Lemma For each gexistential Pformula ψ and guniversal Ptheory Γ {ϕ}: Γ {( x)ϕ( x)} = G ψ Γ {ϕ( t) t U(P)} = G ψ. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
93 Herbrand Expansions The PHerbrand expansion E(ϕ) of a Pformula ϕ consists of formulas obtained by applying the following steps repeatedly: (1) Replace ϕ[( x)ψ( x, y)] where ψ is quantifierfree with ϕ[ t H ψ( t, y)] for some finite H U(P). (2) Replace ϕ[( x)ψ( x, y)] where ψ is quantifierfree with ϕ[ t H ψ( t, y)] for some finite H U(P). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
94 Herbrand Expansions The PHerbrand expansion E(ϕ) of a Pformula ϕ consists of formulas obtained by applying the following steps repeatedly: (1) Replace ϕ[( x)ψ( x, y)] where ψ is quantifierfree with ϕ[ t H ψ( t, y)] for some finite H U(P). (2) Replace ϕ[( x)ψ( x, y)] where ψ is quantifierfree with ϕ[ t H ψ( t, y)] for some finite H U(P). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
95 Herbrand Expansions The PHerbrand expansion E(ϕ) of a Pformula ϕ consists of formulas obtained by applying the following steps repeatedly: (1) Replace ϕ[( x)ψ( x, y)] where ψ is quantifierfree with ϕ[ t H ψ( t, y)] for some finite H U(P). (2) Replace ϕ[( x)ψ( x, y)] where ψ is quantifierfree with ϕ[ t H ψ( t, y)] for some finite H U(P). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
96 More General Herbrand Theorems Theorem For every guniversal Ptheory Γ {ϕ} and gexistential Pformula ψ: Γ {ϕ} = G ψ there exists ϕ E(ϕ) such that Γ {ϕ } = G ψ. Theorem For every guniversal Ptheory Γ and gexistential Pformula ψ: Γ = G ψ there exists ψ E(ψ) such that Γ = G ψ. These theorems hold for any finitary K e.g., if K is the class of complete chains of a variety of integral commutative residuated lattices whose class of chains admits regular completions (see also recent work of Terui). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
97 More General Herbrand Theorems Theorem For every guniversal Ptheory Γ {ϕ} and gexistential Pformula ψ: Γ {ϕ} = G ψ there exists ϕ E(ϕ) such that Γ {ϕ } = G ψ. Theorem For every guniversal Ptheory Γ and gexistential Pformula ψ: Γ = G ψ there exists ψ E(ψ) such that Γ = G ψ. These theorems hold for any finitary K e.g., if K is the class of complete chains of a variety of integral commutative residuated lattices whose class of chains admits regular completions (see also recent work of Terui). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
98 More General Herbrand Theorems Theorem For every guniversal Ptheory Γ {ϕ} and gexistential Pformula ψ: Γ {ϕ} = G ψ there exists ϕ E(ϕ) such that Γ {ϕ } = G ψ. Theorem For every guniversal Ptheory Γ and gexistential Pformula ψ: Γ = G ψ there exists ψ E(ψ) such that Γ = G ψ. These theorems hold for any finitary K e.g., if K is the class of complete chains of a variety of integral commutative residuated lattices whose class of chains admits regular completions (see also recent work of Terui). George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
99 Failure of the Herbrand Theorem in Lukasiewicz Logic Easily, by prenexation, So, by Skolemization right, = Ł ( x)( y)(d(x) D(y)). = Ł ( x)(d(f (x)) D(x)). For Herbrand s Theorem, we would need for some n N, n = Ł (D(f i+1 (c)) D(f i (c))), i=1 but we can build a structure M such that D(f i+1 (c)) M > D(f i (c)) M for i = 1... n, where such a disjunction is not valid. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
100 Failure of the Herbrand Theorem in Lukasiewicz Logic Easily, by prenexation, So, by Skolemization right, = Ł ( x)( y)(d(x) D(y)). = Ł ( x)(d(f (x)) D(x)). For Herbrand s Theorem, we would need for some n N, n = Ł (D(f i+1 (c)) D(f i (c))), i=1 but we can build a structure M such that D(f i+1 (c)) M > D(f i (c)) M for i = 1... n, where such a disjunction is not valid. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
101 Failure of the Herbrand Theorem in Lukasiewicz Logic Easily, by prenexation, So, by Skolemization right, = Ł ( x)( y)(d(x) D(y)). = Ł ( x)(d(f (x)) D(x)). For Herbrand s Theorem, we would need for some n N, n = Ł (D(f i+1 (c)) D(f i (c))), i=1 but we can build a structure M such that D(f i+1 (c)) M > D(f i (c)) M for i = 1... n, where such a disjunction is not valid. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
102 Failure of the Herbrand Theorem in Lukasiewicz Logic Easily, by prenexation, So, by Skolemization right, = Ł ( x)( y)(d(x) D(y)). = Ł ( x)(d(f (x)) D(x)). For Herbrand s Theorem, we would need for some n N, n = Ł (D(f i+1 (c)) D(f i (c))), i=1 but we can build a structure M such that D(f i+1 (c)) M > D(f i (c)) M for i = 1... n, where such a disjunction is not valid. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
103 Approximate Consequence Let us define for a Ptheory Γ, Psentence ψ, and r [0, 1] Q, Γ = Ł r < ψ r < ψ M for every model M of Γ, and note that if the propositional atom P does not occur in Γ {ψ}, Γ = Ł n n+1 < ψ Γ {ψ (P Pn )} = Ł. Firstorder Lukasiewicz logic is not finitary, but Γ = Ł Γ = Ł for some finite Γ Γ. Moreover, for every universal theory Γ {ϕ}, Γ {( x)ϕ( x)} = Ł Γ {ϕ( t) t U(P)} = Ł. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
104 Approximate Consequence Let us define for a Ptheory Γ, Psentence ψ, and r [0, 1] Q, Γ = Ł r < ψ r < ψ M for every model M of Γ, and note that if the propositional atom P does not occur in Γ {ψ}, Γ = Ł n n+1 < ψ Γ {ψ (P Pn )} = Ł. Firstorder Lukasiewicz logic is not finitary, but Γ = Ł Γ = Ł for some finite Γ Γ. Moreover, for every universal theory Γ {ϕ}, Γ {( x)ϕ( x)} = Ł Γ {ϕ( t) t U(P)} = Ł. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
105 Approximate Consequence Let us define for a Ptheory Γ, Psentence ψ, and r [0, 1] Q, Γ = Ł r < ψ r < ψ M for every model M of Γ, and note that if the propositional atom P does not occur in Γ {ψ}, Γ = Ł n n+1 < ψ Γ {ψ (P Pn )} = Ł. Firstorder Lukasiewicz logic is not finitary, but Γ = Ł Γ = Ł for some finite Γ Γ. Moreover, for every universal theory Γ {ϕ}, Γ {( x)ϕ( x)} = Ł Γ {ϕ( t) t U(P)} = Ł. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
106 Approximate Consequence Let us define for a Ptheory Γ, Psentence ψ, and r [0, 1] Q, Γ = Ł r < ψ r < ψ M for every model M of Γ, and note that if the propositional atom P does not occur in Γ {ψ}, Γ = Ł n n+1 < ψ Γ {ψ (P Pn )} = Ł. Firstorder Lukasiewicz logic is not finitary, but Γ = Ł Γ = Ł for some finite Γ Γ. Moreover, for every universal theory Γ {ϕ}, Γ {( x)ϕ( x)} = Ł Γ {ϕ( t) t U(P)} = Ł. George Metcalfe (University of Bern) FirstOrder Logics and Truth Degrees LATD / 39
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