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March 10: Quantificational Notions Interpretations: Every interpretation interprets every individual constant, predicate, and sentence letter of PL. Our.

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Presentation on theme: "March 10: Quantificational Notions Interpretations: Every interpretation interprets every individual constant, predicate, and sentence letter of PL. Our."— Presentation transcript:

1 March 10: Quantificational Notions Interpretations: Every interpretation interprets every individual constant, predicate, and sentence letter of PL. Our partial interpretation may not include some predicates or individual constants, but it does interpret them. Lx, Lxy, Lxyz, Lwxyz a, b, c, d, … a a, b, c, d, … a 1, b 2 … Full interpretations are infinitely long as PL includes an infinite # of predicates, an infinite # of individual constants, an infinite # of sentence letters and an infinite # of UDs.

2 March 10: Quantificational Notions We use the concept of an interpretation to specify the quantificational counterparts of truth-functional concepts. Individual sentences of PL fall into 1 of 3 categories: A sentence P of PL isquantificationally true IFF P is true on every interpretation. A sentence P of PL is quantificationally true IFF P is true on every interpretation. A sentence P of PL is quantificationally false IFF P is false on every interpretation. A sentence P is quantificationally indeterminate IFF P is neither quantificationally true nor quantificationally false.

3 March 10: Quantificational Notions A sentence P of PL isquantificationally true IFF P is true on every interpretation. A sentence P of PL is quantificationally true IFF P is true on every interpretation. How could we demonstrate that some sentence P is quantificationally true given that we cannot check all of the infinite interpretations of that sentence? We cannot do it with all sentences, but we can do it for some using reasoning. Consider for example: (  y) (Cy v ~Cy)

4 We may reason as follows: Because the sentence is existentially quantified, it is true on an interpretation just in case at least one member of the UD satisfies the conditions specified by ‘Cy v ~Cy’ – that is, if at least one member of the UD either is C or is not C. Without knowing what the interpretation of C is, we know that every member of a UD satisfies this condition – for every interpretation interprets C, and every member of the UD is or is not in the extension of C. And because, by definition, every interpretation has a nonempty set as its UD, then the UD for any interpretation has at least one member and hence at least one member that satisfies the open sentence. So the sentence is quanticationally true.

5 In general, to show that some sentence P is quantificationally true, we may use reasoning to show that no matter what the UD is and no matter how the individual constants, predicates, and sentence letters are interpreted, the sentence always turns out to be true. So consider: (  x) (  y) Bxy  (  y) (  x) Bxy Whatever the UD and however B is interpreted, we know that the antecedent is either true or false. If it is true, so is the consequent. If the antecedent is false, the sentence is true. So the sentence is quantificationally true.

6 A sentence P is quantificationally false IFF P is false on every interpretation. (  x) Bx & (  z) ~Bz is quantificationally false. If an interpretation makes the left conjunct true, then it makes the right conjunct false. If it makes the left conjunct false, then the sentence is false. As every interpretation includes ‘B’ as a predicate, and however it interprets it, it cannot be the case that every member of the UD is in the extension of B but one member is not.

7 It is not always as easy to demonstrate that a sentence is quantificationally true or that it is quantificationally false. But we can often show that a sentence is not quantificationally true by showing that it is false on at least one interpretation. (  x) (Wx  Mx)  (  x) Mx 1. UD: the set of all living things Wx: x is a whale Mx: x is a mammal The antecedent is true on this interpretation but the consequent is false.

8 We can often show that a sentence is not quantificationally false by showing that it is true on at least one interpretation. (  x) (  y) (Syx  ~Sxy) 2. UD: the set of positive integers Sxy: x is smaller than y IF ‘There is a positive integer x such that all positive integers are smaller than x…’ The antecedent is false on this interpretation (as no positive integer is smaller than 1), so the sentence is true on this interpretation, and so the sentence is not quantificationally false.

9 We can often show that a sentence is not quantificationally false by showing that it is true on at least one interpretation. (  x) Ex & (  x ) ~Ex 3. UD: the set of positive integers Ex: x is even Or consider: ~(~Ga & (  y) Gy) To construct an interpretation on which this sentence is true (and thus not quantificationally false), we must construct an interpretation on which ~Ga & (  y) Gy is false. And that means making one or the other of the conjuncts false.

10 ~(~Ga & (  y) Gy) We must construct an interpretation on which ~Ga & (  y) Gy is false. And that means making one or the other of the conjuncts false. 4. UD: Set of positive integers Gx: x is even a: 2 Here the left conjunct is false and so is the conjunction…

11 Again, we can show a sentence is not quantificationally true, or that it is not quantificationally false, by constructing a single interpretation that demonstrates this. But we cannot construct a single interpretation that will demonstrate that a sentence is quantificationally true or that it is quantificationally false. For some sentences, we can arrive at such a conclusion by reasoning; but for many, we cannot.

12 A sentence P of PL is quantificationally indeterminate IFF P is neither quantificationally true nor quantificationally false. We show that a sentence is quantificationally indeterminate by constructing two interpretations: one on which it is true (to show that it is not quantificationally false) and one on which it is false (to show that it is not quantificationally true).

13 A sentence P of PL is quantificationally indeterminate IFF P is neither quantificationally true not quantificationally false. We showed, using interpretation 4, that the following sentence is not quantificationally false by finding an interpretation on which it is true: ~(~Ga & (  y) Gy). Now we need an interpretation on which it is false to show that it is also not quantificationally true. This means we need an interpretation on which ~Ga & (  y) Gy is true.

14 5. UD: set of positive integers Gx: x is odd Gx: x is odd a: 2 a: 2 On this interpretation, ~Ga is true (for 2 is not odd) and (  y) Gy) is true (for there is at least one odd positive integer). So the conjunction is true, and its negation is false. Interpretations 4 and 5 demonstrate that the sentence ~(~Ga & (  y) Gy is quantificationally indeterminate.

15 Finding an interpretation on which a sentence is true or on which it is false takes some ingenuity. Determine if the sentence is one whose quantificational status can be settled by an interpretation or can be settled by reasoning (or by neither). Guidelines: Examine the kind of sentence it is. If it is a truth-functional compound, use the truth conditions for that kind of compound. If the sentence is universally quantified, then the sentence is true IFF the condition specified after the quantifier is satisfied by all members of the UD you choose.

16 If the sentence is existentially quantified, then it will be true IFF the condition specified after the quantifier is satisfied by at least one member of the UD. Sometimes, the desired interpretation can not be found. For example: If a sentence is quantificationally true, there is no interpretation on which it is false, and any attempt to construct an interpretation on which the sentence is false will fail. The same point holds for quantificationally false sentences. A set of positive integers is always a good choice for your UD as you construct interpretations.

17 Quantificational notions An argument of PL is quantificationally valid IFF there is no interpretation on which every premise is true and the conclusion is false. An argument of PL is quantificationally invalid IFF the argument is not quantificationally valid.

18 (  x) (Fx v Gx) (  x) ~Fx ------------------ (  x) Gx is quantificationally valid. Suppose that on some interpretation both premises are true. If the first premise is true, then some member x of the UD is either F or G. If the second premise is true, then no member of the UD is F. Therefore, because the member that is either F or G is not F, it must be G. Thus (  x) Gx will be true on any such interpretation.

19 Demonstrating that an argument is not guantificationally valid: find an interpretation in which all of its premises are true and its conclusion is false. Consider: (  x) [(  y) Fy  Fx] (  y) ~Fy ----------------------- ~ (  x) Fx We can make the first premise true by interpreting F so that at least one member of the UD is in its extension – for then that object will satisfy the condition specified by (  y) Fy  Fx beause it will satisfy its consequent.

20 (  x) [(  y) Fy  Fx] (  y) ~Fy ----------------------- ~ (  x) Fx The second premise will be true if at least one member of the UD is not in the extension of F. So F will have some, but not all, members of the UD in its extension. And because some members will be in the extension, the conclusion will be false. 6: UD: set of positive integers Fx: x is odd. Fx: x is odd.

21 Again, there are asymmetries in what we can prove. We can prove an argument is valid only on specific interpretations, not all possible interpretations, although some we can prove are valid by reasoning. But we can prove an argument is quantificationally invalid by finding a single interpretation on which it is not valid (as to be quantificationally valid, an argument must be such that there is no interpretation on which all its premises are true and its conclusion is false).

22 Again, there are asymmetries in what we can prove. We can also prove an argument is not quantificationally invalid by finding an interpretation on which it is valid. Quantificational equivalency Sentences P and Q are quantificationally equivalent IFF there is no interpretation on which P and Q have different truth values. The following are quantificationally equivalent: (  x) Fx  Ga (  x) (Fx  Ga)

23 (  x) Fx  Ga (  x) (Fx  Ga) We reason as follows: Suppose that (  x) Fx  Ga is true on some interpretation. Then (  x) Fx is either true or false on this interpretation. If (  x) Fx is true, then so is Ga. But then since Ga is true, every object x in the UD is such that if x is F, then a is G. So (  x) (Fx  Ga) is true.

24 (  x) Fx  Ga (  x) (Fx  Ga) If (  x) Fx is false, then every object x in the UD is such that if x is F (which we assume here it is not), then a is G and the whole sentence is true. Again, ‘(  x) (Fx  Ga) is also true on that interpretation.

25 (  x) Fx  Ga (  x) (Fx  Ga) Now suppose that (  x) Fx  Ga is false on some interpretation. Then (  x) Fx is true and Ga is false. But if (  x) Fx is true, then some object x in the UD is in the extension of F. This object does not satisfy the condition that if it is F (which it is), then a is G (which it is not on our present assumption). So (  x) (Fx  Ga) is false if (  x) Fx  Ga is false. Taken together with the result that if one is true, so is the other, we have demonstrated that the 2 sentences are quantificationally equivalent.

26 Quantificational consistency A set of sentences of PL is quantificationally consistent IFF there is at least one interpretation on which all the members of the set are true. A set of sentences of PL is quantificationally inconsistent IFF the set is not quantificationally consistent. The set {(  x) ~Bax, ~Bba v (  x) ~Bax} is quantificationally consistent.

27 {(  x) ~Bax, ~Bba v (  x) ~Bax} 7. UD: set of possible integers Bxy: x is less than or equal to y Bxy: x is less than or equal to y a: 1 b: 2 On this interpretation (  x) ~Bax is true since 1 is less than or equal to every positive integer. So, too, ~Bba is true since 2 is neither less than nor equal to 1, so ~Bba v (  x) ~Bax is true


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