4.2 First-Order Logic

Transcription

1 64 First-Order Logic and Type Theory The problem can be seen in the two qestionable rles In the existential introdction, the term a has not yet been introdced into the derivation and its se can therefore not be jstified Related is the incorrect application of the E rle It is spposed to introdce a new parameter a and a new assmption w However, a occrs in the conclsion, invalidating this inference In this case, the flaw can be repaired by moving the existential elimination downward, in effect introdcing the parameter into the derivation earlier (when viewed from the perspective of normal proof constrction) x nat A(s(x)) tre a a nat natis w s(a) nat A(s(a)) tre I y nat A(y) tre E a,w y nat A(y) tre I ( x nat A(s(x))) y nat A(y) tre Of corse there are other cases where the flawed rle cannot be repaired For example, it is easy to constrct an incorrect derivation of ( x τ A(x)) x τ A(x) 42 First-Order Logic First-order logic, also called the predicate calcls, is concerned with the stdy of propositions whose qantifiers range over a domain abot which we make no assmptions In or case this means we allow only qantifiers of the form x τ A(x) and x τ A(x) that are parametric in a type τ We assme only that τ type, bt no other property of τ When we add particlar types, sch as natral nmbers nat or lists τ list, we say that we reason within specific theories The theory of natral nmbers, for example, is called arithmetic When we allow essentially arbitrary propositions and types explained via introdction and elimination constrcts (inclding fnction types, prodct types, etc) we say that we reason in type theory It is important that type theory is open-ended: we can always add new propositions and new types and even new jdgment forms, as long as we can explain their meaning satisfactorily On the other hand, first-order logic is essentially closed: when we add new constrcts, we work in other theories or logics that inclde first-order logic, bt we go beyond it in essential ways We have already seen some examples of reasoning in first-order logic in the previos section In this section we investigate the trth of varios other propositions in order to become comfortable with first-order reasoning Jst like propositional logic, first-order logic has both classical and constrctive variants We prse the constrctive or intitionistic point of view We can recover classical trth either via an interpretation sch as Gödel s translation 1, or by adding 1 detailed in a separate note by Jeremy Avigad

2 42 First-Order Logic 65 the law of exclded middle The practical difference at the first-order level is the interpretation of the existential qantifier In classical logic, we can prove a proposition x τ A(x) tre by proving x τ A(x) tre instead Sch a proof may not yield the witness object t sch that A(t) is satisfied, which is reqired nder the constrctive interpretation of the existential qantifier Bt how is it possible to provide witnesses in pre logic, withot any assmptions abot the domain of qantifiers? The answer is that assmptions abot the existence of objects will be introdced locally dring the proof Bt we have to be carefl to verify that the objects we se to witness existential qantifiers or instantiate niversal qantifiers are indeed assmed to exist and are available at the right point in the derivation As a first concrete example, we investigate the interaction between negation and qantification We prove ( x τ A(x)) x τ A(x) tre The sbject of the jdgment above is a proposition, assming τ type and x τ A(x) prop Since all qantifiers range over the same type τ, we will omit the type label from qantification in all propositions below The reader shold keep in mind that this is merely a shorthand Frthermore, we will not explicitly state the assmption abot the propositional or predicate parameters sch as A(x) v c x A(x) c τ w E A(c) A(c) x A(x) E c,w I v x A(x) I ( x A(x)) x A(x) The two-dimensional notation for derivations becomes difficlt to manage for large proofs, so we extend the linear notation from Section 28 We se the following concrete syntax x τ A(x) x τ A(x) c τ!x:t A(x)?x:t A(x) c : t The qantifiers and act like a prefix operator with minimal binding strength, so that x τ A(x) B is the same as x τ (A(x) B) One complication introdced by existential qantification is that the elimination rle introdces two new assmptions, c τ and A(c) tre In order to

3 66 First-Order Logic and Type Theory distingish between inferred and assmed jdgments, new assmptions are separated by commas and terminated by semi-colon Under these conventions, the for rles for qantification take the following form: Introdction Elimination c : t;?x:t A(x); [c : t,?x:t A(x); ; B]; B; [c : t;!x:t A(x); ; c : t; A(c)];!x:t A(x) We se c as a new parameter to distingish parameters more clearly from bond variables Their confsion is a common sorce of error in first-order reasoning And we have the sal assmption that the name chosen for c mst be new (that is, may not occr in A(x) orb) in the existential elimination and niversal introdction rles Below we restate the proof from above in the linear notation [?x:t ~A(x); [!x:t A(x); [ c : t, ~ F ]; F ]; ~!x:t A(x) ]; (?x:t ~A(x)) => ~!x:t A(x); The opposite implication does not hold: even if we know that it is impossible that A(x) is tre for every x, this does not necessarily provide s with enogh information to obtain a witness for x A(x) In order to verify that this cannot be proven withot additional information abot A, we need to extend or notion of normal and netral proof This is straightforward only the existential elimination rle reqires some thoght It is treated in analogy with disjnction Γ,c τ A(c) I Γ x τ A(x) Γ t τ Γ A(t) I Γ x τ A(x) Γ x τ A(x) Γ A(t) Γ t τ E Γ x τ A(x) Γ,c τ, A(c) C E Γ C

4 42 First-Order Logic 67 In the case of pre first-order logic (that is, qantification is allowed only over one nknown type τ), normal proofs remain complete A correspondingly strong property fails for arithmetic, that is, when we allow the type nat This sitation is familiar from mathematics, where we often need to generalize the indction hypothesis in order to prove a theorem This generalization means that the reslting proof does not have a strong normality property We will retrn to this topic in the next section Now we retrn to showing that ( x A(x)) x A(x) tre is not derivable We search for a normal proof, which means the first step in the bottom-p constrction is forced and we are in the state x A(x) x A(x) I ( x A(x)) x A(x) At this point it is impossible to apply the existential introdction rle, becase no witness object of type τ is available So we can only apply the implication elimination rle, which leads s to the following sitation x A(x) x A(x) x A(x) E x A(x) I ( x A(x)) x A(x) Now we can either repeat the negation elimination (which leads nowhere), or se niversal introdction c x A(x) c τ A(c) I c x A(x) x A(x) E x A(x) I ( x A(x)) x A(x) The only applicable rle for constrcting normal dedctions now is again the implication elimination rle, applied to the assmption labelled This leads to

5 68 First-Order Logic and Type Theory the identical sitation, except that we have an additional assmption d τ and try to prove A(d) Clearly, we have made no progress (since the assmption c τ is now seless) Therefore the given proposition has no normal proof and hence, by the completeness of normal proofs, no proof As a second example, we see that ( x A(x)) x A(x) tre does not have a normal proof After one forced step, we have to prove x A(x) x A(x) At this point, no rle is applicable, since we cannot constrct any term of type τ Intitively, this shold make sense: if the type τ is empty, then we cannot prove x τ A(x) since we cannot provide a witness object Since we make no assmptions abot τ, τ may in fact denote an empty type (sch as 0), the above is clearly false In classical first-order logic, the assmption is often made that the domain of qantification is non-empty, in which case the implication above is tre In type theory, we can prove this implication for specific types that are known to be non-empty (sch as nat) We can also model the standard assmption that the domain is non-empty by establishing the corresponding hypothetical jdgment: c τ ( x τ A(x)) x τ A(x) We jst give this simple proof in or linear notation [ c : t; [!x:t A(x);?x:t A(x) ]; (!x:t A(x)) =>?x:t A(x)]; We can also discharge this assmption to verify that y (( x A(x)) x A(x)) tre withot any additional assmption This shows that, in general, y B is not eqivalent to B, evenify does not occr in B! While this may be conterintitive at first, the example above shows why it mst be the case The point is that while y does not occr in the proposition, it does occr in the proof and can therefore not be dropped