A Completeness Theorem for a 3-Valued Semantics for a First-order Language Christopher Gauker July 20, 2015 This document presents a Gentzen-style deductive calculus and proves that it is complete with respect to a 3-valued semantics for a language with quantifiers. The semantics resembles the strong Kleene semantics with respect to conjunction, disjunction and negation. The completeness proof for the sentential fragment fills in the details of a proof sketched in Arnon Avron (2003) "Classical Gentzen-type Methods in Propositional Many-valued Logics" in Beyond Two: Theory and Application of Multiple-Valued Logics, M. Fitting and E. Orlowska, eds., pp. 117-155. Physica Verlag. The extension to quantifiers is original but uses standard techniques. Sentential Logic Let SL be a sentential language with connectives ¬, !, ^, _ and standard syntax. Let GS3 be a Gentzen-style deductive calculus with the following rules: (Basis) A ) A i.e. ; A ) A (Weakening)   )    0,  )  , 0 (Cut)  1 )  1, A A, 2 )  2  1, 2 )  1, 2 (? )) ¬A,A ) i.e. ;¬A,A ) ; (¬¬ )) A,  )  ¬¬A,  )   1 () ¬¬)   )  , A   )  ,¬¬A (!))   )  , A B,  )   A ! B,  )   ()!)  , A )  , B   )  , A ! B (¬ !)) A,¬B,  )  ¬(A ! B),  )   () ¬ !)   )  , A   )  ,¬B   )  ,¬(A ! B) (^ ))  , A,B )    , A ^B )   () ^)   )  , A   )  , B   )  , A ^B (¬^ ))  ,¬A )    ,¬B )    ,¬(A ^B) )   () ¬^)   )  ,¬A,¬B   )  ,¬(A ^B) (_ ))  , A )    , B )    , A _B )   () _)   )  , A,B   )  , A _B (¬_ ))  ,¬A,¬B )    ,¬(A _B) )   () ¬_)   ), ,¬A   )  ,¬B   )  ,¬(A _B) NB: Arbitrary rearrangements of elements before ")" and arbitrary rearrangements of elements after ")" are allowed. Definition. A valuation v is an assignment to atomic sentences of SL of members of {Y, I,N} ("yes", "indeterminate" and "no"). Definition. V extends a valuation v to every sentence of SL i↵ for all sentences P of SL: 1. if P is atomic: V (P ) = v(P ) 2 2. if P = ¬Q, then: (a) V (P ) = Y if V (Q) = N , (b) V (P ) = N if V (Q) = Y , (c) V (P ) = I otherwise; 3. if P = (Q ! R), then: (a) V (P ) = Y if V (Q) 2 {I,N} or V (R) = Y , (b) V (P ) = N if V (Q) = Y and V (R) = N , (c) V (P ) = I if V (Q) = Y and V (R) = I; 4. if P = (Q^R), then: (a) V (P ) = Y if V (Q) = V (R) = Y , (b) V (P ) = N if V (Q) = N or V (R) = N , (c) V (P ) = I otherwise; 5. if P = (Q_R), then: (a) V (P ) = Y if V (Q) = Y or V (R) = Y , (b) V (P ) = N if V (Q) = V (R) = N , (c) V (P ) = I otherwise; In tables: ¬ Y N I I N Y ! Y I N Y Y I N I Y Y Y N Y Y Y ^ Y I N Y Y I N I I I N N N N N _ Y I N Y Y Y Y I Y I I N Y I N NB: This set of connectives is not functionally complete. That is, not all truth functions on {Y, I, N} can be defined by means of them (Avron 2003, p. 219). Definition. A model for a sequence   )   is a valuation v s.t. if V extends v, then for some P 2  , V (P ) 2 {I,N} or for some P 2  , V (P ) = Y . Definition.  1 )  1, 2 )  2, . . . , n )  n   )   is valid3 i↵ for every valuation v, if v is a model of each of  1 )  1, 2 )  2, . . . , n )  n, then it is a model of   )  . 3 Definition.   |=3   i↵ ;   )   is valid. Definition. Where S = { 1 )  1, 2 )  2, . . . , n )  n}, an S-cut is an application of (Cut) in which A 2 ✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆ . Definition. An S-proof of   )   from a set of sequences S is a proof in which every application of (Cut) is an S-cut. Definition. Where S = { 1 )  1, 2 )  2, . . . , n )  n},  ⇤ )  ⇤ is S-saturated i↵: 1. there is no S -proof of  ⇤ )  ⇤; 2. if A 2 ✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆ then A 2  ⇤ [ ⇤; 3. (a) if ¬¬A 2  ⇤, then A 2  ⇤, (b) if ¬¬A 2  ⇤, then A 2  ⇤; 4. (a) if A ! B 2  ⇤, then A 2  ⇤ or B 2  ⇤, (b) if A ! B 2  ⇤, then A 2  ⇤ and B 2  ⇤, (c) if ¬(A ! B) 2  ⇤, then A 2  ⇤ and ¬B 2  ⇤, (d) if ¬(A ! B) 2  ⇤, then A 2  ⇤ or ¬B 2  ⇤; 5. (a) if A ^B 2  ⇤, then A 2  ⇤ and B 2  ⇤, (b) if A ^B 2  ⇤, then A 2  ⇤ or B 2  ⇤, (c) if ¬(A ^B) 2  ⇤, then ¬A 2  ⇤ or ¬B 2  ⇤, (d) if ¬(A ^B) 2  ⇤, then ¬A 2  ⇤ and ¬B 2  ⇤; 6. (a) if A _B 2  ⇤, then A 2  ⇤ or B 2  ⇤, (b) if A _B 2  ⇤, then A 2  ⇤ and B 2  ⇤, (c) if ¬(A _B) 2  ⇤, then ¬A 2  ⇤ and ¬B 2  ⇤, (d) if ¬(A _B) 2  ⇤, then ¬A 2  ⇤ or ¬B 2  ⇤. 4 NB: If  ⇤ )  ⇤ is S -saturated, then membership in  ⇤ behaves like Y and membership in  ⇤ behaves like I or N. Let s1, s2, . . . , sm, . . . be a list of all formulas that are either subformulas or negations of subformulas in (  [ ) [ ✓✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆◆ . Construct  ⇤ )  ⇤ thus: Let  0 =  , 0 =  . For all i   0, let  i+1 =  i [ {s i+1} , i+1 =  i, if there is no S -proof of  i, s i+1 )  i. For all i   0, let  i+1 =  i, i+1 =  i [ {s i+1}, if there is an S -proof of  i, s i+1 )  i. Let  ⇤ = 1S i=1  i and  ⇤ = 1S i=1  i. Observation 1: A 2  ⇤[ ⇤ i↵ A is a subformula or a negation of a subformula of a formula in (  [ ) [ ✓✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆◆ . Lemma 1. Suppose there is no S-proof of   )  . Then: (i) For each i   0,  i+1 )  i+1 has no S-proof. (ii) There is no S-proof of  ⇤ )  ⇤. (iii) Maximality: Let  0 and  0 be sets consisting of formulas that are either subformulas or negations of subformulas of formulas in (  [ )[ ✓✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆◆ . If  0 *  ⇤ or  0 *  ⇤, then  0, ⇤ )  ⇤, 0 has an S-proof. Proof. (i) By induction: Basis: By assumption,  0 )  0 has no S -proof. Induction hypothesis: Suppose  i )  i has no S -proof. Induction step: By the construction, either  i+1 =  i [ {s i+1} or  i+1 =  i[{s i+1}. If  i+1 =  i[{si+1} , then, by the construction,  i+1 )  i+1 has no S -proof. If  i+1 =  i [ {s i+1}, then there is an S -proof of  i, si+1 )  i. Suppose  i+1 )  i+1 , i.e.  i )  i, s i+1, has an S -proof. Then by (Cut),  i )  i has an S -proof, contrary to assumption. (ii) Suppose there is an S -proof of  ⇤ )  ⇤. Since proofs are finite, there is an i such that   i )   i (in our construction) has an S -proof, contrary to (i). (iii) Let  0 and  0 be as described. Case (a):  0 *  ⇤. There is   0 0 ✓  0 such that   0 0 6= ; and   0 0 \  ⇤ = ;. By the construction,   0 0 ✓  ⇤. So by (Weakening), there is an S -proof of  0, ⇤ )  ⇤, 0. Case (b):  0 *  ⇤. Similarly. 5 Lemma 2. If   )   has no S-proof, then  ⇤ )  ⇤ (as in the construction) is S-saturated. Proof. Suppose   )   has no S-proof. Condition (1) in the definition of S -saturated: By Lemma 1(ii). Condition (2) in the definition of S -saturated: Suppose, for a reductio, that A 2 ✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆ but A /2  ⇤ [ ⇤. By Lemma 1, maximality, A, ⇤ )  ⇤ and  ⇤ )  ⇤, A have S -proofs. But   ⇤ )  ⇤, A A, ⇤ )  ⇤  ⇤ )  ⇤ is an application of (Cut) to a member of ✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆ . So  ⇤ )  ⇤ has an S -proof, contrary to Lemma 1(ii). Condition (3): (a) Suppose ¬¬A 2  ⇤. A,  ⇤ )  ⇤ ¬¬A, ⇤ )  ⇤ is an application of (¬¬ )). But ¬¬A, ⇤ =  ⇤. So  ⇤ )  ⇤ has an S -proof if A, ⇤ )  ⇤ has one. So A, ⇤ )  ⇤ has no S -proof. So A /2  ⇤. So by the construction (and Observation 1), A 2  ⇤. (b) Suppose ¬¬A 2  ⇤. By () ¬¬),  ⇤ )  ⇤ has an S -proof if  ⇤ )  ⇤, A has one. So  ⇤ )  ⇤, A has none. So A /2  ⇤. So A 2  ⇤. Condition (4): (a) Suppose A ! B 2  ⇤.   ⇤ )  ⇤, A B, ⇤ )  ⇤ A ! B, ⇤ )  ⇤ is an applcation of (!)). But A ! B, ⇤ =  ⇤. Since  ⇤ )  ⇤ lacks an S -proof, either (i)  ⇤ )  ⇤, A has no S -proof, or (ii) B, ⇤ )  ⇤ has no S -proof. Suppose (i). A /2  ⇤. So by the construction, A 2  ⇤. Suppose (ii). B /2  ⇤. So B 2  ⇤. (b) Suppose A ! B 2  ⇤. A,  ⇤ )  ⇤, B  ⇤ )  ⇤, A ! B is an application of ()!). But  ⇤, A ! B =  ⇤. Since  ⇤ )  ⇤ has no S -proof  ⇤, A )  ⇤, B has no S -proof. So A /2  ⇤, B /2  ⇤, which means A 2  ⇤ and B 2  ⇤. (c) Suppose ¬(A ! B) 2  ⇤. A,¬B,  ⇤ )  ⇤ ¬(A ! B), ⇤ )  ⇤ is an application of (¬ !)). But ¬(A ! B), ⇤ =  ⇤. So since  ⇤ )  ⇤ has no S -proof, A,¬B, ⇤ )  ⇤ has no S -proof. So A /2  ⇤,¬B /2  ⇤, so A 2  ⇤,¬B 2  ⇤. (d) Suppose ¬(A ! B) 2  ⇤.   ⇤ )  ⇤, A  ⇤ )  ⇤,¬B  ⇤ )  ⇤,¬(A ! B) is an application of () ¬ !). But  ⇤,¬(A ! B) =  ⇤. Since  ⇤ )  ⇤ has no S -proof, either (i)  ⇤ )  ⇤, A has no S -proof, or (ii)  ⇤ )  ⇤,¬B has no S -proof. So either A /2  ⇤ or ¬B /2  ⇤. So either A 2  ⇤ or ¬B 2  ⇤. Conditions (5) and (6): Similarly. Lemma 3. If  ⇤ )  ⇤, constructed from   )  , is S-saturated , then there is a valuation that is a model of every sequence in S, but not a model of   )  . 6 Proof. Suppose  ⇤ )  ⇤ is S -saturated. Define valuation v as follows: For all atomic formulas P of SL, v(P ) = 8 >>>< >>>: Y if P 2  ⇤ I if P /2  ⇤ and¬P /2  ⇤ N if ¬P 2  ⇤ . First step: v is well-defined: If P 2   and ¬P 2  ⇤, then by (Weakening) P,¬P )  ⇤ )  ⇤ will be an S -proof. Since   ⇤ )  ⇤ (by S-saturation) does not have an S -proof, either P /2  ⇤ or ¬P /2  ⇤. So the definition of v does not yield both v(P ) = Y and v(P ) = N . Second step: Suppose V extends v. Prove for all formulas P of SL, if P 2  ⇤ then V (P ) = Y and if P 2  ⇤ then V (P ) 2 {I,N}. By induction: Basis: The thesis holds for all literals (atomic sentences and negations of atomic sentences): First, consider atomic P. (i) Suppose P 2  ⇤. v(P ) = V (P ) = Y . (ii) Suppose P 2  ⇤. Since  ⇤ )  ⇤ has no S -proof, P /2  ⇤. If ¬P /2  ⇤, then v(P ) = V (P ) = I 2 {I,N}. If ¬P 2  ⇤, then v(P ) = V (P ) = N 2 {I,N}. Next, consider ¬P , where P is atomic. (i) Suppose ¬P 2  ⇤. v(P ) = V (P ) = N . V (¬P ) = Y . (ii) Suppose ¬P 2  ⇤. Since  ⇤ )  ⇤ has no S -proof, ¬P /2  ⇤. If P /2  ⇤, then v(P ) = V (P ) = I and V (¬P ) = I 2 {I,N}. If P 2  ⇤, then v(P ) = V (P ) = Y and V (¬P ) = N 2 {I,N}. Induction hypothesis: The thesis holds for A,B,¬A and ¬B. Induction step: Show that it holds for ¬¬A, (A ! B),¬(A ! B), (A ^ B),¬(A ^B), (A _B) and ¬(A _B). (¬¬): Suppose ¬¬A 2  ⇤. By the definition of S -saturation, A 2  ⇤. By IH, V (A) = Y . V (¬¬A) = Y . Suppose ¬¬A 2  ⇤. By the definition of S -saturation, A 2  ⇤. By IH, V (A) 2 {I,N}. V (¬¬A) 2 {I,N}. (!): Suppose (A ! B) 2  ⇤. By the definition of S -saturation, A 2  ⇤ or B 2  ⇤. By IH, V (A) 2 {I,N}, or V (B) = Y . V (A ! B) = Y . Suppose (A ! B) 2  ⇤. By the definition of S -saturation, A 2  ⇤ and B 2  ⇤. By IH, V (A) = Y, V (B) 2 {I,N}. V (A ! B) 2 {I,N}. (¬ !) Suppose ¬(A ! B) 2  ⇤. By the definition of S -saturation, A 2  ⇤ and ¬B 2  ⇤. By IH, V (A) = Y and V (¬B) = Y , V (¬(A ! B)) = Y . Suppose ¬(A ! B) 2  ⇤. By the definition of S -saturation, A 2  ⇤ or ¬B 2  ⇤. By IH, V (A) 2 {I,N} or V (¬B) 2 {I,N}. V (A) 2 {I,N} or V (B) 2 {Y, I}. V (A ! B) 2 {Y, I}. V (¬(A ! B)) 2 {I,N}. Cases (^), (¬^), (_), (¬_) similarly. Consequently, v is not a model of  ⇤ )  ⇤. So since (by Observation 1) 7   ✓  ⇤ and   ✓  ⇤, v is not a model of   )  . Third step: Show that v is a model of every sequence in S. Let   i )   i be an arbitrary member of S. Show that v is a model of   i )   i . Suppose, for reductio, that   i ✓  ⇤ and  i ✓  ⇤. In that case, by (Weakening),  ⇤ )  ⇤ has an S -proof,   i )   i  ⇤ )  ⇤ , contrary to Lemma 1(ii). So, either  i *   ⇤ or   i *  ⇤. But by condition (2) in the definition of S -saturation (or by Observation 1),   i [  i ✓  ⇤ [ ⇤. So either (i) there is A 2   i such that A 2  ⇤, or (ii) there is A 2   i such that A 2  ⇤. In case (i) V (A) 2 {I,N}. So v is a model of   i )   i . In case (ii) V (A) = Y . So v is a model of   i )   i . Completeness Theorem for GS3: If  1 )  1, 2 )  2, . . . , n )  n   )   is valid3, then, where S = { 1 )  1, 2 )  2, . . . , n )  n}, there is an S-proof of   )   in GM3. Proof. Suppose there is no S -proof of   )  . Then, by Lemma 2,   )   can be extended to S -saturated  ⇤ )  ⇤. By Lemma 3,  1 )  1, 2 )  2, . . . , n )  n   )   is not valid3. Corollary 1. If   |=3   then there is a proof in GS3 of   )  . Extension of these results to QL Suppose that SL is now a language like SL, defined above, except that the atomic formulas are composed, by the usual syntax, from countably many predicates of each adicity and denumerably many variables and denumerably many individual constants. Let QL be a language containing, for each (individual) variable v, a quantifier 8v. QL has the standard syntax, allowing vacuous quantification, and 9v abbreviates ¬8v¬. In any sequence,   )  , the members of   [  are sentences, not open formulas. Pn/v denotes the result of substituting n for v wherever v occurs free in P. If v is not in P, then Pn/v = P . Let GQ3 be a Gentzen-style deductive calculus containing all of the rules of GS3 plus the following: (8 )) Pn/v,  )  8vP,  )   () 8)   )  , Pn/v   )  , 8vP where n is not in P and not in any member of  [ , i.e. n is new, or v is not in P, 8 (¬8 )) ¬Pn/v,  )  ¬8vP,  )   where n is not in P and not in any member of  [ , i.e. n is new, or v is not in P, () ¬8)   )  ,¬Pn/v   )  ,¬8vP Define a 3-valued structure M as a triple hU,⌃+,⌃ i where U, the universe, is a set of objects, and for each individual constant n, ⌃+(n) = ⌃ (n) = ⌃(n) 2 U . For each m-ary predicate R, ⌃+(R) ✓ Um,⌃ (R) ✓ Um, and ⌃+(R)\⌃ (R) = ;. Let a structure and variable assignment M g be a quadruple hU,⌃+,⌃ , gi with U,⌃+,⌃  as before and g a partial function over some of the variables of QL such that for each variable v in the range of g : g(v) 2 U . g[v/o] is a variable assigment like g except that g[v/o] assigns o to v instead of whatever g assigned to v, if v is in the range of g, and otherwise assigns o to v, if v is in the range of g. g; is the empty variable assignment with an empty range. Associate with M g the function h such that for each singular term t of QL that is either an individual constant of QL or a variable of QL in the range of g, h(t) = 8 < : ⌃(t) if t is an individual constant, g(t) if t is a variable. A structure M = M g; . Associate with each structure and variable assignment M g a function of the same name from formulas of QL into {Y, I,N}, as follows: M g (Rt1t2 . . . tm) = Y i↵ hh(t1), h(t2), . . . , h(tm)i 2 ⌃+(R), M g (Rt1t2 . . . tm) = N i↵ hh(t1), h(t2), . . . , h(tm)i 2 ⌃ (R), M g (Rt1t2 . . . tm) = I otherwise, M g (¬P ) = Y i↵ M g (P ) = N , M g (¬P ) = N i↵ M g (P ) = Y , M g (¬P ) = I otherwise, M g (P ! Q) = Y i↵ M g (P ) 2 {I,N} or M g (Q) = Y , M g (P ! Q) = N i↵ M g (P ) = Y and M g (Q) = N , M g (P ! Q) = I otherwise, M g (8vQ) = Y i↵ for all o 2 U , M g[v/o](Q) = Y , M g (8vQ) = N i↵ for some o 2 U , M g[v/o](Q) = N , M g (8vQ) = I otherwise. 9 Observation 2: If U is identical to the set of all individual constants of QL and for all singular constants n of QL, ⌃(n) = n, then M(8vQ) = Y i↵ for all n of QL, M(Qn/v) = Y , and M(8vQ) = N i↵ for some n of QL, M(Qn/v) = N . Observation 3: If v is not in Q, M g (8vQ) = Y i↵ M g (Q) = Y and M g (8vQ) = N i↵ M g (Q) = N . Definition. A structure M is a model for   )   i↵ either there is P 2   such that M(P ) 2 {I,N} or there is P 2   such that M(P ) = Y . Definition.  1 )  1, 2 )  2, . . . , n )  n   )   is validQ3 i↵ every structure that is a model for each of  1 )  1, 2 )  2, . . . , n )  n is a model for   )  . Let C be a denumerable set of individual constants not in QL. QL+ is QL supplemented by the individual constants in C. By standard techniques, we associate with each formula P of QL+ having exactly one free variable, two members of C, c+ P and c  P , called the witnesses for P, having the same birth date, such that for no formula Q of QL+ whose witnesses have that same birth date or an earlier birth date does Q contain c+ P or c  P . c+ P is the positive witness for P and c  P is the negative witness for P. Definition. The Henkin set H for QL+ is the set of sentences Q of QL+ such that Q 2 H i↵ 1. n is an individual constant of QL+ and Q = (8vP ! Pn/v) or Q = (¬Pn/v ! ¬8vP ), or 2. v is not in P, and Q = (P ! 8vP ) or Q = (¬8vP ! ¬P ), or 3. c+ P is the positive witness for P and Q = (Pc+ P /v ! 8vP ), or 4. c  P is the negative witness for P and Q = (¬8vP ! ¬Pc  P /v). Lemma 4. (a) (A ! B),  )     )  , A is provable. (b) (A ! B),  )    , B )   is provable. Proof. (a) A ) A  , A )  , B,A (Weakening)   )  , A, (A ! B) ()!) (A ! B),  )     )   (Cut) 10 (b) Similarly. Lemma 5.  , (8vP ! Pn/v) )     )   is provable.  , (Pn/v ! 8vP ) )     )   where n is new, or v is not in P, is provable.  , (¬8vP ! ¬Pn/v) )     )   where n is new, or v is not in P, is provable.  , (¬Pn/v ! ¬8vP ) )     )   is provable. Proof. By Lemma 4 and (8 )), () 8), (¬8 )) and () ¬8) respectively. For example:  , (¬8vP ! ¬Pn/v) )     )  ,¬8vP (Lemma 4 )  , (¬8vP ! ¬Pn/v) )   ¬Pn/v ,  )   (Lemma 4 ) ¬8vP ,  )   (¬8 ))   )   (Cut) The Elimination Theorem Suppose every sentence in (  [ ) [ ✓✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆◆ is in QL. Suppose also that  1,H )  1, . . . , n,H )  n  ,H )   is provable in GQ3 for QL +. Then  1 )  1, . . . , n )  n   )   is provable in GQ3 for QL. Proof. Suppose the hypothesis. Since proofs are finite, there is a finite subset J ✓ H such that  1,J )  1, . . . , n,J )  n  ,J )   is provable. Show:  1 )  1, . . . , n )  n   )   is provable (in GQ3 for QL). By induction on the size of J : Basis: J = ;. Trivial. Induction Hypothesis : Suppose the thesis holds when J hast m members (m   0). Show that the thesis holds when J has m+ 1 members. Case 1: At least one member Q of J is of the form (8vP ! Pn/v) or (¬Pn/v ! ¬8vP ) or (P ! 8vP ) or (¬8vP ! P ). There is a set J 0 such that J = J 0 [ {Q}, Q /2 J 0. By the IH it su ces to show that if  1,J 0 [ {Q} )  1, . . . , n,J 0 [ {Q} )  n  ,J 0 [ {Q} )   is provable then 11  1,J 0 )  1, . . . , n,J 0 )  n  ,J 0 )   is provable, thus:  1,J 0 )  1  1,J 0 [ {Q} )  1 (Weakening) . . . . . .   n ,J 0 )   n   n ,J 0 [ {Q} )   n (Weakening)  ,J 0 [ {Q} )   (Supposition)  ,J 0 )   (by Lemma 5) . Case 2 : All members of J are sentences of the form (Pc+ P /v ! 8vP ) or (¬8vP ! ¬Pc  P /v). Of all the witnesses in the sentences in J (positive or negative), let c⇤ P be one that has latest birth date. (There might be two, one positive, one negative.) c⇤ P does not occur in any member of (  [ ) [✓✓ nS i=1   i ◆ [ ✓ nS i=1   i ◆◆ or in any other member of J . Let Q be a sentence in J containing c⇤ P . So Q = (Pc⇤ P /v ! 8vP ) or Q = (¬8vP ! ¬Pc⇤ P /v). As in Case 1, we can drop Q from the proof. Definition. A valuation val for QL+ is an assignment of the members of {Y, I,N} to sentences of QL⇤ that are either quantified or atomic. Definition. Val extends val to every sentence of QL+ in accordance with tables given above for SL. The Henkin Construction Theorem: Suppose val is a valuation for QL+ and Val extends val such that for all Q 2 H, V al(Q) = Y . Then we can construct a structure M V al for QL+ such that for all sentences P of QL+, V al(P ) = Y i↵ M V al (P ) = Y , and V al(P ) = N i↵ M V al (P ) = N (By implication: V al(P ) = I i↵ M V al (P ) = I). Proof. Define M V al thus: U is identical to the set of individual constants of QL+. For each individual constant n of QL+: ⌃(n) = n. For each m-place predicate R of QL+(QL): ⌃+(R) = {hn1, n2, . . . , nmi|val(Rn1n2 . . . nm) = Y }, ⌃ (R) = {hn1, n2, . . . , nmi|val(Rn1n2 . . . nm) = N}. By induction on the length of sentences: Basis: Suppose P is atomic, i.e. P = Rn1n2 . . . nm. Left-to-right : Suppose V al(Rn1n2 . . . nm) = Y . By the construction of M V al , hn1, n2, . . . , nmi 2 ⌃+(R). By the construction of MV al, h⌃(n1),⌃(n2), . . . ,⌃(n m )i 2 ⌃+(R). So by the definition ofM g (as a function),M V al (Rn1n2 . . . nm) = Y. Suppose V al(Rn1n2 . . . nm) = N . By the construction, hn1, n2, . . . , nmi 2 12 ⌃ (R). By the construction, h⌃(n1),⌃(n2), . . . ,⌃(nm)i 2 ⌃+(R). So M V al (Rn1n2 . . . nm) = N . Right-to-left : Suppose M V al (Rn1n2 . . . nm) = Y . By the construction of M V al , h⌃(n1),⌃(n2), . . . ,⌃(nm)i 2 ⌃+(R). By the construction of MV al, hn1, n2, . . . , nmi2 ⌃+(R). V al(Rn1n2 . . . nm) = Y . SupposeMV al(Rn1n2 . . . nm) = N. Similarly. Induction hypotheses: Suppose the thesis holds for all sentences having complexity k. Show that it holds for all sentences having complexity k+1. Induction step: (¬): Exercise. (!): Suppose P = (Q ! R). Left-to-right : Suppose V al(Q ! R) = Y . By the definition of Val, either V al(Q) 2 {I,N} or V al(R) = Y . By IH, eitherM V al (Q) 2 {I,N} orM V al (R) = Y . By the definition of M g ,M V al (Q ! R) = Y . Suppose V al(Q ! R) = N . By the definition on Val, V al(Q) = Y and V al(R) = N . By the definition of M g ,M V al (Q ! R) = N . Right-to-left : Exercise. (^), (_): Exercise. (8): Left-to-right: Suppose V al(8vQ) = Y . By the definition of H, for all individual constants n in QL+, V al(8vQ ! Qn/v) = Y. So by the definition of Val, for all individual constants n in QL+, V al(Qn/v) = Y . By the IH, for all individual constants n in QL+, M V al (Qn/v) = Y . By Observation 2, M V al (8vQ) = Y . Suppose V al(8vQ) = N . V al(¬8vQ) = Y . Case 1 : v is not in Q. Then by the construction of H, V al(¬8vQ ! ¬Q) = Y . By the definition of V al, V al(¬Q) = Y. By IH, M V al (¬Q) = Y . By the definition of M g ,M V al (Q) = N . By Observation 3, M V al (8vQ) = N . Case 2 : v is in Q. Then by the construction of H, V al(¬8vQ ! ¬Qc  Q /v) = Y . By the definition of V al, V al(¬Qc  Q /v) = Y . By the IH, M V al (¬Qc  Q /v) = Y . By the definition of M g ,M V al (Qc  Q /v) = N . By the definition of M g ,M V al (8vQ) = N . Right-to-left : SupposeM V al (8vQ) = Y . Case 1: v is not inQ. By Observation 3, M V al (Q) = Y . By IH, V al(Q) = Y . By the construction of H, V al(Q ! 8vQ) = Y . V al(8vQ) = Y . Case 2 : v is in Q. By Observation 2, for all individual constants n in QL+,M V al (Qn/v) = Y . In particular, M V al (Qc+ Q /n) =Y. By IH, V al(Qc+ Q /v) = Y . By the construction of H, V al(Qc+ Q /n ! 8vQ) = Y . V al(8vQ) = Y . Suppose M V al (8vQ) = N. By Observation 2, there is an individual constant n of QL+ such that M V al (Qn/v) = N . By IH, V al(Qn/v) = N . V al(¬Qn/v) = Y . By the construction of H, 13 V al(¬Qn/v ! ¬8vP ) = Y . V al(¬8vP ) = Y . V al(8vP ) = N . Completeness Theorem for GQ3: If  1 )  1, . . . , n )  n   )   is validQ3, then  1 )  1, . . . , n )  n   )   is provable in GQ3. Proof. Suppose that  1 )  1, . . . , n )  n   )   is not provable in GQ3. By the Elimination Theorem,  1,H )  1, . . . , n,H )  n  ,H )   is also not provable in GQ3. So it is also not provable in GS3. By the Completeness Theorem for GS3,  1,H )  1, . . . , n,H )  n  ,H )   is not valid3. So there is a valuation val such that val is a model for each of  1,H )  1, . . . , n,H )  n, but not a model for  ,H )  . Since val is not a model for  ,H )  , for all Q 2 H, V al(Q) = Y . By the Henkin Construction Theorem, there is a structure M V al , such that for all Q 2 H, M V al (Q) = Y , but M V al is a model for each of  1,)  1, . . . , n,)  n, but not for   )  . So  1 )  1, . . . , n )  n   )   is not valid Q3.