Download Lecture 11 What did we do so far ? • We defined first order structures

Sets and Logic (234293) WS 2014/5
Lecture 11
Lecture 11
What did we do so far ?
• We defined first order structures and showed examples.
• We introduced the syntax of first order logic F OL.
• We started introducing the semantics of first order logic.
• Meaning function for terms.
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Lecture 11
Outline of lecture 11
• We continue the semantics of first order logic.
• Meaning function for atomic formulas.
• Meaning function for formulas.
• Satisfiability, validity and logical consequence.
• Definability in first order logic.
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Definition 1 (4.3.19)
Semantics of First Order Logic
The semantic for our languages is described by means of the
meaning function
M (φ, A, z)
where φ is a formula, A a structure and z an assignment.
The meaning function is defined inductively, using the Unique
Readability Theorem for First Order Logic.
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Definition 1 contd (4.3.19)
Semantics of First Order Logic (Contd)
Basis:
equality:
M ((t1 ≈ t2), A, z) = 1
iff
MT(t1, z) = MT(t2, z);
relation:
M (Rj,α(t1 . . . , tj ), A, z) = 1
iff
(MT(t1, z), . . . , MT(tj , z)) ∈ A(Rj,α);
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Definition 1 contd (4.3.19)
Semantics of First Order Logic (Contd)
Closure (Boolean connectives):
¬: M (¬φ, A, z) = TT¬ (M (φ, A, z))
∨: M ((φ ∨ ψ), A, z) =
TT∨ (M (φ, A, z), M (ψ, A, z));
∧: M ((φ ∧ ψ), A, z) =
TT∧ (M (φ, A, z), M (ψ, A, z));
→: M ((φ → ψ), A, z) =
TT→ (M (φ, A, z), M (ψ, A, z));
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Definition 1 contd (4.3.19)
Semantics of First Order Logic (Contd)
Closure (quantifiers):
∃: M ((∃vi φ), A, z) = 1 iff
there exists z1 ∈ Ass(A) such that
z =i z1 and M (φ, A, z1 ) = 1;
∀: M ((∀vi φ), A, z) = 1 iff
for every z1 ∈ Ass(A) such that
z =i z1 we have that M (φ, A, z1 ) = 1
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Examples
φ(v1 , v2 , v3 ) = ”v12 + v22 = v32 ” :
(F+ (F× (v1 , v1 ), F× (v2 , v2 )) ≈ F× (v3 , v3 ))
ψ(v2 ) = ”there are v1 , v3 with v12 + v22 = v32 ” :
∃v1 ∃v3 (F+ (F× (v1 , v1 ), F× (v2 , v2 )) ≈ F× (v3 , v3 ))
We look at two assignments z and z 0 :
z(v1 ) = 5, z(v2 ) = 12, z(v3 ) = 13
z 0 (v1 ) = 5, z 0 (v2 ) = 4, z 0 (v3 ) = 3
We compute the meaning function in the structure N:
M (φ, N, z) = 1,
M (ψ, N, z) = 1,
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M (φ, N, z 0 ) = 0
M (ψ, N, z 0 ) = 1
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Examples
φ1 = ∀v1 ∃v2 R< (v1 , v2 )
φ2 = ∀v2 ∃v1 R< (v1 , v2 )
φ3 = ∃v1 ∀v2 R< (v1 , v2 )
φ4 = ∃v1 ∀v2 (R< (v1 , v2 ) ∨ (v1 ≈ v2 ))
Our assignment is
z(v1 ) = 3, z(v2 ) = 2, z(v3 ) = 1,
We compute the meaning function in N and Z,
with (m, n) ∈ N(R< ) iff m < n and
with (m, n) ∈ Z(R< ) iff m < n.
M (φ1 , N, z) = 1
M (φ2 , N, z) = 0
M (φ3 , N, z) = 0
M (φ4 , N, z) = 1
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M (φ1 , Z, z) = 1
M (φ2 , Z, z) = 1
M (φ3 , Z, z) = 0
M (φ4 , Z, z) = 0
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n is a prime number
We construct formulas
• φdiv (v1 , v2 ) says:
v1 divides v2 properly.
∃v3 ((F× (v1 , v3 ) ≈ v2 ) ∧
(¬(v3 ≈ c1 ) ∧ ¬(v3 ≈ v2 )))
• φprime(v2 ) says:
v2 is a prime number ( v2 is a variable, n is a number):
¬∃v1 φdiv (v1 , v2 )
which, fully written, is
¬∃v1 ∃v3 ((F× (v1 , v3 ) ≈ v2 ) ∧
(¬(v3 ≈ c1 ) ∧ ¬(v3 ≈ v2 )))
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There are infinitely many prime numbers
We construct formulas
• φinf (ψ(v2 )) says:
There are arbitrarily large v2
such that ψ(v2 ).
∀v1 ∃v2 (R< (v1 , v2 ) ∧ ψ(v2 ))
• φinf −primes says:
there are arbitrarily large primes:
∀v1 ∃v2 (R< (v1 , v2 ) ∧ φprime(v2 ))
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Checking M (φ, A, z) = 1
To show that M (φinf −primes, N, z) = 1, we have to show that there are
arbitrarily large primes in N.
This needs a proof.
Given z(v1 ) = m, let p1 , p2 , . . . , pk(m) be
all the prime numbers smaller or equal to m.
Put m1 = p1 · p2 · . . . · pk(m) + 1 and let q(m) be the smallest prime dividing m1 .
We find that q ≥ pi for every i ≤ k(m),
and q ≥ m.
We put now z(v2 ) = q(m).
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gcd(a, b) = c
We build the formula φgcd(v1, v2, v3)
which says that gcd(v1, v2) = v3
• First we define ψ(v1, v2, v3),
which says that v3 divides both v1 and v2.
(∃v4 (F×(v1, v4) ≈ v3) ∧ ∃v4 (F×(v2, v4) ≈ v3))
• Now we put φgcd(v1, v2, v3) =
(ψ(v1, v2, v3)∧ ∀v4 (ψ(v1, v2, v4) →
(R<(v4, v3) ∨ (v4 ≈ v3)))
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There are infinitely many prime numbers p(x) = a · x + b,
provided gcd(a, b) = 1.
(Dirichlet’s Theorem, 1837)
We use the formula φinf −lin =
∀v1∀v3∀v4 φgcd(v3, v4, c1) →
∃v2 R<(v1, v2) ∧ φprime F+(v3, F×(v4, v2))
To see that M (φinf −lin, N, z) = 1 we use
Dirichlet’s theorem, which says exactly this.
The proof is highly non-trivial.
Paulo Ribenboim, The Book of Prime Number Records, Springer
1988
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There are infinitely many prime number twins
A pair (p, q) of prime numbers is a twin if q = p + 2.
We use the formula φinf −twins =
∀v1∃v2
R<(v1, v2) ∧ φprime(v2)
∧φprime F+(v2, F+(c1, c1))
To show that M (φinf −twins, N, z) = 1 we have to show that there
are arbitrarily large prime twins in N.
This is an open problem in Number Theory.
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Example 2 (Exercise 4.5.7)
The FOL-Induction Principle
Here we give a FOL-version of the Induction principle.
Let Ind+φ be the following formula:
∀v1, . . . , ∀vk [(φ(c0, v1, . . . , vk )∧
∀v0 φ(v0, v1, . . . , vk ) → φ(F+(v0, c1), v1, . . . , vk ))
→ ∀v0φ(v0, v1, . . . , vk )]
Verify that N |= Ind+φ for every τarith-formula φ.
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Conclusion
• Checking M (φ, A, z) = 1
in a particular structure A and for a particular formula φ
is Mathematics
• Studying properties of the meaning function M (φ, A, z)
is Logic
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Basic Semantic Concepts of First Order Logic
As for Propositional Logic we now define for First Order Logic:
• satisfiability,
• tautologies,
• logical equivalence and
• logical consequence.
As particular examples and applications of the notion of logical equivalence
we study the effect of substitution of subformulas and some normal forms for
first order logic.
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Definition 3 (4.4.1)
Satisfiability and (Logical) Validity,
Let Σ be a set of τ -formulas and φ and ψ be two τ -formulas.
(i) We extend the meaning function to sets of formulas:
M (Σ, A, z) = 1 iff for every φ ∈ Σ we have that M (φ, A, z) = 1.
(ii) Σ is satisfiable iff there is a τ -structure A and an assignment z such that
M (Σ, A, z) = 1
(iii) φ is a tautology (or logically valid) iff
for every τ -structure A and
for every assignment z we have that
M (Σ, A, z) = 1
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Examples 4
The following are tautologies (logically valid):
Let φ(v1 ), ψ(v1 ) and θ(v1 , v2 ) be a F OL-formulas with free variables v1 , v2 as
indicated.
∃v1 (R1,0 (v1 ) ∨ ¬R1,0 (v1 ))
∃v1 (φ(v1 ) ∨ ¬φ(v1 ))
∀v2 (v2 ≈ v2 )
(φ(v1 ) ∨ ¬φ(v1 ))
∀v1 (φ(v1 ) ∨ ¬φ(v1 ))
(φ(v1 ) → (ψ(v1 ) → φ(v1 )))
∃v1 (φ(v1 ) → (ψ(v1 ) → φ(v1 )))
(∀v1 ∀v2 θ(v1 , v2 ) → ∀v2 ∀v1 θ(v1 , v2 ))
(∃v1 ∃v2 θ(v1 , v2 ) → ∃v2 ∃v1 θ(v1 , v2 ))
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Definition 5 (4.4.1, contd)
Logical Equivalence and Logical Consequence
Let Σ be a set of τ -formulas and φ and ψ be two τ -formulas.
Let A be a τ -structure and z be an assignment.
(i) φ is a logical consequence of Σ,
and we write Σ |= φ,
if for every τ -structure A and every assignment z such that
M (Σ, A, z) = 1 we have also that M (φ, A, z) = 1.
(ii) We say that φ and ψ are logically equivalent,
and we write φ ≡ ψ,
if {φ} |= ψ and if {ψ} |= φ.
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Notation
(i) If Σ = ∅
we write simply
|= φ instead of ∅ |= φ.
(ii) If Σ = {ψ} is a singleton we write,
by abuse of notation,
also ψ |= φ instead of {ψ} |= φ.
(iii) Similarly, we write
Σ |= Σ1 if
Σ |= φ for every φ ∈ Σ1.
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Proposition 6 (4.4.3)
As for propositional logic
(i) φ is a tautology iff ¬φ is not satisfiable.
(ii) φ is a tautology (or valid) if it is a consequence of the empty set, i.e.
|= φ.
(iii) Σ |= (φ → ψ) iff Σ ∪ {φ} |= ψ.
(iv) Σ |= (φ ∧ ¬φ) iff Σ is not satisfiable.
(v) Σ is not satisfiable iff for every φ we have that Σ |= φ.
(vi) φ ≡ ψ iff both |= (φ → ψ) and |= (ψ → φ)
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Proposition 7 (4.4.4 (i))
Tautologies obtained by substitution
Let B(p1 , . . . , pn ) ∈ WFF be a propositional formula
which is a (propositional) tautology.
Let φ1 , . . . , φn be τ -formulas.
Let φ be the τ -formula
B(φ1 , . . . , φn )
obtained from B by replacing each occurrence
of the propositional variable pi by φi .
Then φ is a tautology (among first order formulas).
Proof: (Exercise)
Compute the meaning function.
First we prove by induction on the structure of B that M (φ, A, z) = MP L(B, z1 )
where z1 is the propositional assignment defined by z1 (pi ) = M (φi , A, z).
Then we use that B is a propositional tautology to conclude that φ is a first
order tautology.
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Proposition 8 (4.4.4 (ii))
Logical equivalences obtained by substitution
Let φ1 , ψ1 and ψ2 be τ -formulas such that ψ1 is a subformula of φ1 and ψ1 ≡ ψ2 .
Let φ2 be the formula obtained from φ1 by replacing ψ1 by ψ2 .
Then φ1 ≡ φ2 .
Proof: (Exercise)
Compute the meaning function.
Let φ1 = α ◦ ψ1 ◦ β.
We proceed by backward induction on α
and induction on β
and use the Unique Readability Theorem.
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Manipulating quantifiers
The most important issue in understanding the
Meaning Function of
First Order Logic
is understanding the
manipulation of quantifiers.
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Proposition 9
Moving quantifiers into formulas
Let φ, ψ be τ -formulas. Then
(i) ∀vi (φ ∧ ψ) ≡ (∀vi φ ∧ ∀vi ψ);
∃vi (φ ∨ ψ) ≡ (∃vi φ ∨ ∃vi ψ);
(ii) If vi does not occur free in ψ then
(∀vi φ ∧ ψ) ≡ ∀vi (φ ∧ ψ) ;
(∀vi φ ∨ ψ) ≡ ∀vi (φ ∨ ψ);
and
(∃vi φ ∨ ψ) ≡ ∃vi (φ ∨ ψ):
(∃vi φ ∧ ψ) ≡ ∃vi (φ ∧ ψ);
(iii) ¬∀vi φ ≡ ∃vi ¬φ
and
¬∃vi φ ≡ ∀vi ¬φ;
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Homework
Study examples of formulas where
vi does occur free in ψ and
show that proposition 9(ii) is false, i.e.,
(∀viφ ∧ ψ) 6≡ ∀vi(φ ∧ ψ) ;
(∀viφ ∨ ψ) 6≡ ∀vi(φ ∨ ψ);
and
(∃viφ ∨ ψ) 6≡ ∃vi(φ ∨ ψ);
(∃viφ ∧ ψ) 6≡ ∃vi(φ ∧ ψ).
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Definition 10 (4.4.6)
Substitution of variables by terms
Let s : Var 7→ T erm(τ ) be a function.
Let t ∈ T erm(τ ) and φ be a τ -formula.
We define first st(t, s) which applies the substitution s to terms:
(i) st(cα , s) = cα ;
(ii) st(vi , s) = s(vi );
(iii) st(Fn,α (t1 , . . . , tn ), s) =
Fn,α (st(t1 , s), . . . , st(tn , s));
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Definition 10 (4.4.6. contd)
Substitution in formulas
(iv) subst((t1 ≈ t2 ), s) = st(t1 , s) ≈ st(t2 , s);
(v) subst(Rn,α (t1 , . . . , tn ), s) =
Rn,α (st(t1 , s), . . . , st(tn , s));
(vi) subst(¬φ, s) = ¬subst(φ, s);
(vii) subst((φ ∧ ψ), s) = (subst(φ, s) ∧ subst(ψ, s));
And similarly for ∨ and →.
(viii) subst(∀vi φ, s) = ∀vi subst(φ, s1 ) with
s1 (vi ) = vi and s1 (vj ) = s(vj ) for all i 6= j.
(ix) subst(∃vi φ, s) = ∃vi subst(φ, s1 ) with
s1 (vi ) = vi and s1 (vj ) = s(vj ) for all i 6= j.
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Proposition 11 (4.4.7)
Renaming of Bound Variables
Let s : Var 7→ Var be a function and φ a τ -formula. Assume that s(vj ) does
not occur at all in φ and s(vi ) = vi for every i 6= j. Then
∀vj φ ≡ ∀s(vj )subst(φ, s)
and
∃vj φ ≡ ∃s(vj )subst(φ, s).
Proof:
The proof uses the fact that M (φ, A, z) is not dependent on z(s(vj )). Q.E.D.
The condition that s(vj ) does not occur at all in φ
in proposition 11 can be weakend further.
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Talking First Order Logic
We now will express
some mathematical concepts
from the Real Numbers
in First Order Logic.
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Dense Linear Orders
Let Σdense−order consist of the following five statements:
Three cases:
∀v1 ∀v2 ((R< (v1 , v2 ) ∨ R< (v2 , v1 )) ∨ v1 ≈ v2 )
Asymmetry:
∀v1 ∀v2 (¬ (R< (v1 , v2 ) ∧ R< (v2 , v1 )))
Transitivity:
∀v1 ∀v2 ∀v3 ((R< (v1 , v2 ) ∧ R< (v2 , v3 ))
→ R< (v1 , v3 ))
Density:
∀v1 ∀v2 (R< (v1 , v2 )
→ ∃v3 (R< (v1 , v3 ) ∧ R< (v3 , v2 )))
There are at least two elements:
∃v1 ∃v2 R< (v1 , v2 )
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Logical Consequences of Σdense−order
For which of the following formulas φ do we have Σdense−order |= φ ?
First element: (no)
∃v1 ∀v2 (¬v1 ≈ v2 → R< (v1 , v2 ))
There are at least three elements: (yes)
∃v1 ∃v2 ∃v3 ((¬v1 ≈ v2 ∧ ¬v2 ≈ v3 ) ∧ ¬v1 ≈ v3 )
Two between each two: (yes)
∀v1 ∀v2 (R< (v1 , v2 ) →
∃v3 ∃v4 (R< (v1 , v3 ) ∧ (R< (v3 , v4 ) ∧ R< (v4 , v2 ))))
No last element: (no)
∀v1 ∃v2 R< (v1 , v2 )
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Ordered Fields, I
Recall τarith = {F+ , F∗ , R< , c0 , c1 } and consider the following statements:
Commutativity:
x + y = y + x, x ∗ y = y ∗ x;
Associativity:
x + (y + z) = (x + y) + z
x ∗ (y ∗ z) = (x ∗ y) ∗ z;
Neutral Element:
0 + x = x, 1 ∗ x = x, 0 ∗ x = 0;
Inverse Element:
For every x there is a y such that x + y = 0,
for every x 6= 0 there is a y such that x ∗ y = 1;
Distributivity:
x ∗ (y + z) = (x ∗ y) + (x ∗ z).
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Ordered Fields, II
Asymmetry:
If x < y then not y < x;
Transitivity:
if x < y and y < z then x < z;
Linearity:
for every x, y either x < y or y < x or x = y;
Monotonicity:
0<1
If x < y then for every z also x + z < y + z;
if 0 < x and 0 < y then also 0 < x ∗ y.
Let Φord−f ield be the conjunction of the translations
of these statements into F OL(τarith ).
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A non-trivial unsatisfiable set of formulas
In the vocabulary of arithmetic
∃v0 (F+ (F∗ (v0 , v0 ), c1 ) ≈ c0 )
says that −1 has a square root.
We look at the statement
In an ordered field
there is no square root of −1.
Formalize this as
Φord−f ield ∪ {∃v0 (F+ (F∗ (v0 , v0 ), c1 ) ≈ c0 )}
is not satisfiable.
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Theorem 12
Φord−f ield ∪ {∃v0(F+(F∗(v0, v0), c1) ≈ c0)} is not satisfiable.
Proof: Let
Σ1 = Φord−f ield ∪ {∃v0 (F+ (F∗ (v0 , v0 ), c1 ) ≈ c0 )}
and
Σ2 (v0 ) = (Φord−f ield ∪ {F+ (F∗ (v0 , v0 ), c1 ) ≈ c0 )}
Assume for contradiction, there is A and z such that
M (Σ1 , A, z) = 1
Let z =0 z 0 with z 0 (v0 ) = a and
M (Σ2 , A, z 0 ) = 1
In other words Σ1 is satisfiable iff Σ2 is satisfiable.
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Proof of Theorem 12 (contd)
Now in A we have either a < 0, a = 0 or 0 < a.
If a = 0, so a2 = 0 = −1 but 0 6= −1.
If 0 < a then 0 < a2 = −1 but −1 < 0.
If a < 0 then
−a > 0 and (−1)a > 0 and (−1)2 a2 = −1 > 0, but −1 < 0.
Each case gives a contradiction.
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Q.E.D.
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Proof sequences for FOL
We shall see a
formalization of proof rules for FOL
at the end of the course.
The system we shall exhibit is sound and complete.
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Lecture 11
Prenex Normal Form
• First order formulas are built by using the boolean operations and quantification in arbitrary order.
• The purpose of Normal Forms is to restrict the order of these building
steps, similar as in the case of CNF for propositional logic.
• The most important normal form for first order logic is the Prenex
Normal Form.
The word ”prenex” refers to ”pre” (latin: before) and ”nexus” (latin: bound)
and is used in logic to indicate that all the variable binding appears in the
formula before any other logical operations.
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Definition 13 (4.4.10)
Quantifier free formulas
The set QF(τ ) of quantifier free τ -formulas is built inductively
as follows:
Basis:
Atomic τ -formulas are quantifier free.
Closure:
If φ, ψ ∈ QF(τ ) then so are
¬φ, (φ ∧ ψ), (φ ∨ ψ), (φ → ψ).
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Definition 14 (4.4.11)
Prenex Normal Form
The set PNF(τ ) of τ -formulas in prenex normal form is built
inductively as follows:
Basis:
QF(τ ) ⊆ PNF(τ ).
Closure:
If φ ∈ PNF(τ ) then so are ∀viφ and ∃viφ.
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Examples for PNF
We look at the formula:
∀v0 (∀v1 R(v1 , v0 ) → ∃v1 R(v0 , v1 ))
We replace → and rename the last bound occurrence of v1 :
∀v0 (¬∀v1 R(v1 , v0 ) ∨ ∃v2 R(v0 , v2 ))
Now we move ¬ inside:
∀v0 (∃v1 ¬R(v1 , v0 ) ∨ ∃v2 R(v0 , v2 ))
Now we move the quantifiers out:
∀v0 ∃v1 ∃v2 (¬R(v1 , v0 ) ∨ R(v0 , v2 ))
and reintroduce →:
∀v0 ∃v1 ∃v2 (R(v1 , v0 ) → R(v0 , v2 ))
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Examples for PNF (contd)
The following are pairs of equivalent formulas, where one is in PNF:
Density:
∀v1 ∀v2 (R< (v1 , v2 ) → ∃v3 (R< (v1 , v3 ) ∧ R< (v3 , v2 )))
∀v1 ∀v2 ∃v3 (R< (v1 , v2 ) → (R< (v1 , v3 ) ∧ R< (v3 , v2 )))
Transitivity:
∀v1 ∀v2 ∀v3 ((R< (v1 , v2 ) ∧ R< (v2 , v3 )) → R< (v1 , v3 ))
∀v1 ∀v3 (∃v2 (R< (v1 , v2 ) ∧ R< (v2 , v3 )) → R< (v1 , v3 ))
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Lecture 11
Examples for PNF (contd)
Consider the conjunction of
First element:
∃v1 ∀v2 (¬v1 ≈ v2 → R< (v1 , v2 ))
No last element:
∀v1 ∃v2 R< (v1 , v2 )
Conjunction:
(∃v1 ∀v2 (¬v1 ≈ v2 → R< (v1 , v2 )) ∧ ∀v1 ∃v2 R< (v1 , v2 ))
in PNF:
∃v1 ∀v2 ∀v3 ∃v4 ((¬v1 ≈ v2 → R< (v1 , v2 )) ∧ R< (v3 , v4 ))
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Sets and Logic (234293) WS 2014/5
Lecture 11
Examples for PNF (contd)
We write now R instead of R< .
φ(v0 ) says there are at least three elements smaller than v0 but uses only
two variables:
∃v1 (R(v1 , v0 ) ∧ ∃v0 (R(v1 , v0 ) ∧ ∃v1 R(v1 , v0 )))
In PNF this looks:
∃v1 ∃v2 ∃v3 ((R(v1 , v0 ) ∧ R(v2 , v1 )) ∧ R(v3 , v2 ))
How many variables do we need to say:
There are exactly 17 elements smaller than v0 ?
Can we do with three variables? In PNF?
v0 is smallest such that ψ(v0 ):
(ψ(v0 ) ∧ ∀v1 (ψ(v1 ) → (R(v0 , v1 ) ∨ v0 ≈ v1 )))
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Sets and Logic (234293) WS 2014/5
Lecture 11
Theorem 15 (4.4.12)
Prenex Normal Form Theorem
For every τ -formula φ there is a τ -formula ψ such that:
ψ ∈ PNF(τ ),
φ ≡ ψ and
φ and ψ have the same free variables.
Furthermore the length of ψ is linear in the length of φ.
Proof:
The proof is by induction on φ.
If φ ∈ PNF(τ ) there is nothing to prove.
In the other cases we use
the proposition on moving quantifiers 9,
renaming bound variables 11
and replacing equivalent subformulas 7. and 8.
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Q.E.D.
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