Introduction to Logic
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Where are we? 1. Preliminaries
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2. Propositional Logic 3. Tree method for PL
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4. First-order Predicate Logic
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5. Tree Method for FOL 6. Expressiveness of FOL
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5. Trees for FOL
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The tree method for predicate logic We add to the branching rules for propositional connectives three rules for quantifiers. We use an infinite set of individual constants {a0 , a1 , ...}. a.
b.
¬∀xφ
¬∃xφ
∃x¬φ
∀x¬φ
∀xφ(x) φ(b)
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∃xφ(x)
: for every constant b already present on the branch. If there is no previous constant, use a0 . :for c a fresh constant, not already present on the tree
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Algorithm 1. To show whether Γ |= φ, (with Γ finite). Append ¬φ to the formulas of Γ. 2. Close any branch which contains a formula and its negation. If all branches are closed, stop. 3. If a formula has not been considered and has for main connective a propositional connective, apply the connective rule. Otherwise do 4. 4. If a formula ∃xφ(x) has not been considered, apply rule and go back to 3. Otherwise go to 5. 5. If a formula ∀xφ(x) has not been examined, apply the rule and go to 6. 6. Whenever a new formula or new constant appears on the tree, go back to stage 2. Otherwise, stop. EALING 2005
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Example 1 “If everyone is proud, then someone is proud” `? ∀xP (x) → ∃xP (x)
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¬(∀xP (x) → ∃xP (x))
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¬(∀xP (x) → ∃xP (x))
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∀xP (x) ¬∃xP (x)
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¬(∀xP (x) → ∃xP (x)) ∀xP (x) ¬∃xP (x)
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∀x¬P (x)
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¬(∀xP (x) → ∃xP (x)) ∀xP (x) ¬∃xP (x)
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√
∀x¬P (x)
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¬(∀xP (x) → ∃xP (x)) ∀xP (x) ¬∃xP (x)
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√
∀x¬P (x) P (a0 )
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¬(∀xP (x) → ∃xP (x)) ∀xP (x) ¬∃xP (x)
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∀x¬P (x) P (a0 )
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¬(∀xP (x) → ∃xP (x)) ∀xP (x) ¬∃xP (x)
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∀x¬P (x) P (a0 ) ¬P (a0 )
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¬(∀xP (x) → ∃xP (x)) ∀xP (x) ¬∃xP (x)
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∀x¬P (x) P (a0 ) ¬P (a0 ) × • The algorithm stops and all branches are closed.
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Example 2 “someone is cooking and reading, and someone is cooking and watching TV, so someone is reading and watching TV” ∃x(A(x) ∧ B(x)), ∃x(A(x) ∧ C(x)) `? ∃x(B(x) ∧ C(x)) ∃x(A(x) ∧ B(x)) ∃x(A(x) ∧ C(x)) ¬∃x(B(x) ∧ C(x))
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∃x(A(x) ∧ B(x)) ∃x(A(x) ∧ C(x)) ¬∃x(B(x) ∧ C(x))
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∀x¬(B(x) ∧ C(x))
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∃x(A(x) ∧ C(x)) ¬∃x(B(x) ∧ C(x))
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∀x¬(B(x) ∧ C(x)) A(a0 ) ∧ B(a0 )
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√ √
¬∃x(B(x) ∧ C(x))
√
∀x¬(B(x) ∧ C(x)) A(a0 ) ∧ B(a0 ) A(a1 ) ∧ C(a1 )
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∃x(A(x) ∧ B(x)) ∃x(A(x) ∧ C(x))
√ √
¬∃x(B(x) ∧ C(x))
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∀x¬(B(x) ∧ C(x)) √ A(a0 ) ∧ B(a0 ) A(a1 ) ∧ C(a1 ) A(a0 ) B(a0 )
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∃x(A(x) ∧ B(x)) ∃x(A(x) ∧ C(x))
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¬∃x(B(x) ∧ C(x))
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∀x¬(B(x) ∧ C(x)) √ A(a0 ) ∧ B(a0 ) √ A(a1 ) ∧ C(a1 ) A(a0 ) B(a0 ) A(a1 ) C(a1 )
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∃x(A(x) ∧ B(x)) ∃x(A(x) ∧ C(x))
√ √
¬∃x(B(x) ∧ C(x))
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∀x¬(B(x) ∧ C(x)) √ A(a0 ) ∧ B(a0 ) √ A(a1 ) ∧ C(a1 ) A(a0 ) B(a0 ) A(a1 ) C(a1 )
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∀x¬(B(x) ∧ C(x)) √ A(a0 ) ∧ B(a0 ) √ A(a1 ) ∧ C(a1 ) A(a0 ) B(a0 ) A(a1 ) C(a1 ) ¬(B(a0 ) ∧ C(a0 ))
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∀x¬(B(x) ∧ C(x)) √ A(a0 ) ∧ B(a0 ) √ A(a1 ) ∧ C(a1 ) A(a0 ) B(a0 ) A(a1 ) C(a1 ) ¬(B(a0 ) ∧ C(a0 )) ¬(B(a1 ) ∧ C(a1 ))
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∀x¬(B(x) ∧ C(x)) √ A(a0 ) ∧ B(a0 ) √ A(a1 ) ∧ C(a1 ) A(a0 ) B(a0 ) A(a1 ) C(a1 ) ¬(B(a0 ) ∧ C(a0 ))
√
¬(B(a1 ) ∧ C(a1 )) ¬B(a0 )
¬C(a0 )
×
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A(a0 ) ∧ B(a0 ) A(a1 ) ∧ C(a1 )
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A(a0 ) B(a0 ) A(a1 ) C(a1 ) ¬(B(a0 ) ∧ C(a0 )) ¬(B(a1 ) ∧ C(a1 )) ¬B(a0 )
√ √
¬C(a0 )
× ¬B(a1 )
¬C(a1 ) ×
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• The algorithm stops, but there is an open branch. • Note that the open branch gives us a countermodel with two individuals a0 and a1 , in which a0 is A, B but not C, and a1 is A, C but not B. U = {a0 , a1 } AM = U B M = {a0 } C M = {a1 }
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Example 3 “Everyone loves someone, so someone is loved by everyone” ∀x∃yL(x, y) `? ∃y∀xL(x, y) ∀x∃yL(x, y) ¬∃y∀xL(x, y) ∀y¬∀xL(x, y) ∀y∃x¬L(x, y)
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Example 3 “Everyone loves someone, so someone is loved by everyone” ∀x∃yL(x, y) `? ∃y∀xL(x, y) ∀x∃yL(x, y) ¬∃y∀xL(x, y) ∀y¬∀xL(x, y) ∀y∃x¬L(x, y)
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∃yL(a0 , y) ∃x¬L(x, a0 ) L(a0 , a1 ) ¬L(a2 , a0 ) ∃yL(a1 , y) ∃yL(a2 , y) ∃x¬L(x, a1 ) ∃x¬L(x, a2 ) L(a1 , a3 ) L(a2 , a4 ) ¬L(a5 , a1 ) ¬L(a6 , a2 ) .. . EALING 2005
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• The algorithm will continue indefinitely : no stop, and an infinite open branch. • However: it is us who can say (and prove) “the branch will be infinite” ; the algorithm has no way to tell. • This also suggests that, from the algorithm, we can construct an infinite model that satisfies ∀x∃yL(x, y), but falsifies ∃y∀xL(x, y).
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Summary 1. The procedure terminates with all branches closed 2. The procedure terminates with some finite open branch 3. The procedure does not terminate (there is an infinite open branch).
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Completeness Let us write : ` φ for : “the tree built from ¬φ closes”. We can prove that : Theorem (Completeness) : ` φ iff |= φ • (soundness) If a branch extension rule is applied to a satisfiable branch, one of the resulting branches is satisfiable. The interesting cases are the rules for ∀ and ∃. Clearly, if a branch containing ∀xφ(x) is satisfiable, so is the extension. If a branch containing ∃xφ(x) is satisfiable in a model M , then for some constant c, φ(c) is satisfied in M . What about φ(c) where c is a fresh constant? Then 0 M0 the model M in which c = cM is also a model of the branch.
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• (completeness) If B is an open branch, extract a model MB by letting UB be the set of constants occurring on that branch. For P an n-ary predicate letter, set: (ai1 , ..., ain ) is in P M iff P (ai1 , ..., ain ) is a formula on B By induction, one can prove that if φ is on B, then MB |= φ, and if ¬φ is on B, then MB |= ¬φ. • So, if 0 φ, then MB , where B is an open branch, satisfies ¬φ, and so 2 φ.
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Undecidability • So, FOL is complete: the tree method yields a sound and complete proof procedure for validity in FOL. But is FOL decidable? • Note that the tree method is not a decision method, but only a semi-decision method: if |= φ, then ` φ, ie the tree will eventually close. But if 2 φ, it can happen that the algorithm keeps on searching (remember example 3!). • This is not a specificity of this method, indeed: Theorem (Church 1936) : FOL is undecidable.
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6. Limitations of FOL
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Expressiveness • We know that FOL is more expressive than PL, but how expressive is it? • To answer this question, we need to know some general properties about FOL. • The compactness theorem, in particular, plays an essential role.
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First-order logic with identity Consider FOL to which one adds a binary relation = expressing identity. In particular, if c and d are constants of the language, and M is a structure, then M |= c = d iff cM = dM . How much can one express with identity? Abbreviation: x 6= y for ¬(x = y) “there are at least two elements” : ∃x∃y(¬x = y) “there are at least three elements”: ∃x∃y∃z(x 6= y ∧ y 6= z ∧ z 6= x) V “there are at least n elements” λn := ∃x1 ∃x2 . . . ∃xn ( xi 6= xj ) i6=j
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“there are at most two elements”: ∀x∀y∀z(x = y ∨ y = z ∨ x = z) W “there are at most n elements”: µn = ∀x0 ∀x1 ∀xn ( xi = xj ) i6=j
“there is exactly n elements” : there are at least n elements, and at most n elements: (λn ∧ µn )
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Some stars are blue : ∃x(S(x) ∧ B(x)) (At least) two stars are blue : ∃x∃y(6= x = y ∧ S(x) ∧ S(y) ∧ B(x) ∧ B(y)) Can we express that: “infinitely many stars are blue”? • Consider the following theory (set of sentences): T = {λn ; n ≥ 2} “there is at least two elements” “there is at least three elements” .. . • T has only infinite models. Suppose it has a finite model M . Let k be the size of the domain. Since M satisfies all sentences of T , M |= λk+1 : contradiction. EALING 2005
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A finite number of Existential quantification: some Finite cardinality quantification : at least n, at most n, exactly n. Infinite cardinality: infinitely many (but note that we expressed it by means of an infinite set of sentences). What about “there is a finite number of”? This can’t be expressed in FOL ! Why? because FOL has compactness.
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Compactness Consider the sentences: F : “there are finitely many elements” F1 : “there is at least one element” F2 : “there are at least two elements” F3 : “there are at least three elements” .. . Taken together, these sentences are inconsistent: they require both that there are finitely many elements, and they entail that there are infinitely many elements. However, every finite subset of these sentences is satisfiable (why?). EALING 2005
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Compactness • We know that each Fn is expressible in FOL by the sentence λn . If F (finiteness) were expressible by a sentence φ, then the set of sentences Γ = {λn ; n ≥ 2} ∪ {φ} would not be satisfiable, while every finite subset would be : so FOL would not be compact. However... • Theorem (Compactness) If every finite set of FO-sentences of a set Γ has a model, then Γ has a model. • Corollary : finiteness is not expressible in FOL.
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Further examples • Some properties of relations can be expressed in FOL: R is reflexive : ∀xR(x, x) R is symmetric : ∀x∀y(R(x, y) → R(y, x)) • Others can’t be expressed. Example : Connectedness: “for any two points x and y, there is a path leading from x to y”, ie a set of points a1 , ..., an such that xRa1 Ra2 R...Ran Ry. Proof: show that the theory saying “x and y are connected, are distinct, and there is no path of length 1 between them, no path of length 2, ...”, is consistent (use compactness).
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Most (1) There are more cats than dogs (2) Most dogs are nice Neither of the sentences can be expressed within FOL. Why not? The idea is: they involve quantification over sets, not only individuals. “there are more A than B”: the set of A has more elements than the set of B. “Most A are B” means : the set of A that are B has more elements than the set of A that are not B.
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Restricted quantification Some A are B : [∃x : A(x)]B(x) All A are B : [∀x : A(x)]B(x) We can define these new expressions within FOL. [∃x : A(x)]B(x) ! ∃x(A(x) ∧ B(x)) [∀x : A(x)]B(x) ! ∀x(A(x) → B(x)]
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Most again Introduce a new quantifier M x : “for most x in the domain” M xP (x) : “most individuals are P ” [M x : A(x)]B(x) is true in a structure if there are more individuals that are A and B than are A and ¬B. Imagine we work with only finite structures: • Show that [M x : A(x)]B(x) cannot be paraphrased as: (a) M x(A(x) → B(x)) (b) M x(A(x) ∧ B(x)) • Show that (a) gives a weaker reading, and that (b) gives a stronger reading. EALING 2005
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Beyond FOL How to go beyond these expressive limitations? Use a more powerful logic: like Second Order Logic (SOL), in which we can quantify over properties (sets) ! Problem : we lose some nice properties. Second-order logic, for instance, is not only undecidable, but even incomplete. It is also non-compact (since it can express finiteness of a domain).
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