Functions Num : N N and Sub : N 3 N , Num( a ) := a , Sub( , - PowerPoint PPT Presentation

Reminder of Num , Sub , Thm N Functions Num : N → N and Sub : N 3 → N , Num( a ) := � a � , Sub( � ϕ � , � x � , � t � ) := � ϕ x t � , defined by ∆-formulas Num ( x, y ) and Sub ( x 1 , x 2 , x 3 , y ). The set Thm N ⊆ N , Thm N := { � ϕ � : N ⊢ ϕ } , is defined by a Σ-formula Thm N ( x ): Thm N ( x ) := ∃ d Deduction N ( d, x ) where Deduction N ( d, x ) is a ∆-formula expressing “ d = � � δ 1 � , . . . , � δ n � � where ( δ 1 , . . . , δ n ) is a deduction from N and x = � δ n � ”.

Representable ⇒ Σ -Definable Previously, we showed that every ∆ -definable set is representable . (This is a straightforward corollary of Proposition 5.3.13: N proves every Σ - sentence which is true in N .) Next, we show that every representable set is Σ -definable .

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . (Without loss of generality, we take v 1 to be the free variable of ϕ .)

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . Let β ( x ) be the following Σ-formula: � � β ( x ) : ≡ ∃ x ∃ y Num ( x, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Thm N ( z ) .

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . Let β ( x ) be the following Σ-formula: � � β ( x ) : ≡ ∃ x ∃ y Num ( x, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Thm N ( z ) . CLAIM: β ( x ) defines A . That is, for every n ∈ N , we have n ∈ A ⇐ ⇒ N | = β ( n ) .

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . Let β ( x ) be the following Σ-formula: � � β ( x ) : ≡ ∃ x ∃ y Num ( x, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Thm N ( z ) . CLAIM: β ( x ) defines A . That is, for every n ∈ N , we have n ∈ A ⇐ ⇒ N | = β ( n ) . First, assume n ∈ A . Since ϕ represents A , we have N ⊢ ϕ ( n ). Therefore, � ϕ ( n ) � ∈ Thm N (by definition of the set Thm N ). Since the formula Thm N ( z ) defines Thm N , it follows that N | = Thm N ( � ϕ ( n ) � ). We have N | = ( Num ( x, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Thm N ( z ))[ s ] under a vari- able assignment function s : Vars → N with s ( x ) = n, s ( y ) = Num( n ) = � n � , s ( z ) = � Sub( � ϕ � , � v 1 � , � n � ) � = � ϕ ( n ) � . Therefore, N | = β ( x )[ s ] ≡ β ( n ).

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . Let β ( x ) be the following Σ-formula: � � β ( x ) : ≡ ∃ x ∃ y Num ( x, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Thm N ( z ) . CLAIM: β ( x ) defines A . That is, for every n ∈ N , we have n ∈ A ⇐ ⇒ N | = β ( n ) . For the other direction, assume N | = β ( n ). Then there exists s : Vars → N with s ( x ) = n such that N | = ( Num ( x, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Thm N ( z ))[ s ]. Since N | = Num ( x, y )[ s ], it must be that s ( y ) = Num( s ( x )) = Num( n ) = � n � . Since N | = Sub ( � ϕ � , � v 1 � , y, z )[ s ], it must be that s ( z ) = � Sub( � ϕ � , � v 1 � , s ( y )) � = � ϕ ( n ) � . Since N | = Thm N ( z )[ s ], we have N | = Thm N ( s ( z )) ≡ Thm N ( � ϕ ( n ) � ). Since Thm N ( z ) defines Thm N , it follows that � ϕ ( n ) � ∈ Thm N , hence N ⊢ ϕ ( n ). Finally, since ϕ represents A , we conclude that n ∈ A . Q.E.D.

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . Let β ( x ) be the following Σ-formula: β ( x ) : ≡ ∃ d “ Deduction N ( d, Sub( � ϕ � , � v 1 � , Num( x )))” . CLAIM: β ( x ) defines A . That is, n ∈ A ⇐ ⇒ N | = β ( n ) for all n ∈ N . First, suppose n ∈ A . Then N ⊢ ϕ ( n ) (since ϕ represents A ). So there exists a deduction ( δ 1 , . . . , δ k ) of ϕ ( n ) from N . Let d := � � δ 1 � , . . . , � δ k � � . Then = “ Deduction N ( d, � ϕ v 1 N | n � )” ≡ “ Deduction N ( d, Sub( � ϕ � , � v 1 � , Num( a ))” � � ≡ ∃ y ∃ z Num ( n, y ) ∧ Sub ( � ϕ � , � v 1 � , y, z ) ∧ Deduction N ( d, z ) . Therefore, N | = β ( n ).

Proposition 6.3.3. If A ⊆ N k is representable, then A is Σ -definable. Proof. Let A ⊆ N for simplicity and assume that ϕ ( v 1 ) represents A . Let β ( x ) be the following Σ-formula: β ( x ) : ≡ ∃ d “ Deduction N ( d, Sub( � ϕ � , � v 1 � , Num( x )))” . CLAIM: β ( x ) defines A . That is, n ∈ A ⇐ ⇒ N | = β ( n ) for all n ∈ N . Conversely, suppose N | = β ( n ), that is, N | = “ Deduction N ( d, � ϕ ( v 1 ) � )” . Then there exists a deduction of ϕ ( v 1 ) from N , hence N ⊢ ϕ ( n ). Since ϕ represents A , it follows that n ∈ A . Q.E.D.

G¨ odel’s Self-Reference Lemma Lemma 6.2.2. Let β ( x ) be an L NT -formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β ( � θ � ) .

G¨ odel’s Self-Reference Lemma Lemma 6.2.2. Let β ( x ) be an L NT -formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β ( � θ � ) . Proof Idea. Let’s first define a formula θ with the weaker property that N | = θ ↔ β ( � θ � ) . We obtain θ by first defining a formula γ ( v 1 ) such that, for every formula ϕ ( v 1 ), N | = γ ( � ϕ � ) ↔ β ( � ϕ ( � ϕ � ) � ) . Once we have such a formula γ ( v 1 ), we simply let θ : ≡ γ ( � γ � ).

G¨ odel’s Self-Reference Lemma Lemma 6.2.2. Let β ( x ) be an L NT -formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β ( � θ � ) . Proof Idea. Let’s first define a formula θ with the weaker property that N | = θ ↔ β ( � θ � ) . We obtain θ by first defining a formula γ ( v 1 ) such that, for every formula ϕ ( v 1 ), N | = γ ( � ϕ � ) ↔ β ( � ϕ ( � ϕ � ) � ) . Definition: � � γ ( v 1 ) : ≡ ( ∃ y )( ∃ z ) Num ( v 1 , y ) ∧ Sub ( v 1 , 8 , y, z ) ∧ β ( z ) (Note that � v 1 � = � 2 � = 2 3 = 8, so 8 ≡ � v 1 � ≡ SSSSSSSS 0.)

G¨ odel’s Self-Reference Lemma Lemma 6.2.2. Let β ( x ) be an L NT -formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β ( � θ � ) . Proof Idea. Let � � γ ( v 1 ) : ≡ ( ∃ y )( ∃ z ) Num ( v 1 , y ) ∧ Sub ( v 1 , � v 1 � , y, z ) ∧ β ( z ) � � θ : ≡ γ ( � γ � ) : ≡ ( ∃ y )( ∃ z ) Num ( � γ � , y ) ∧ Sub ( � γ � , � v 1 � , y, z ) ∧ β ( z ) .

G¨ odel’s Self-Reference Lemma Lemma 6.2.2. Let β ( x ) be an L NT -formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β ( � θ � ) . Proof Idea. So, we have � � γ ( v 1 ) : ≡ ( ∃ y )( ∃ z ) Num ( v 1 , y ) ∧ Sub ( v 1 , � v 1 � , y, z ) ∧ β ( z ) � � θ : ≡ γ ( � γ � ) : ≡ ( ∃ y )( ∃ z ) Num ( � γ � , y ) ∧ Sub ( � γ � , � v 1 � , y, z ) ∧ β ( z ) . � �� � forces variable assignment y �→ �� γ �� We see that forces variable assignment z �→ � γ ( � γ � ) � (= � θ � ) � �� � � � N | = θ ⇐ ⇒ N | = ( ∃ z ) Sub ( � γ � , � v 1 � , �� γ �� , z ) ∧ β ( z ) ⇐ ⇒ N | = β ( � θ � ) .

G¨ odel’s Self-Reference Lemma Lemma 6.2.2. Let β ( x ) be an L NT -formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β ( � θ � ) . Proof Idea, continued. This formula θ has the property that N | = θ ↔ β ( � θ � ) . In order to prove the stronger assertion N ⊢ θ ↔ β ( � θ � ) , we need to replace ∆-formulas Num ( x, y ) and Sub ( x 1 , x 2 , x 3 , y ) in θ with: Num ∗ ( x, y ) : ≡ Num ( x, y ) ∧ ( ∀ z < y )[ ¬ Num ( x, z )] Sub ∗ ( x 1 , x 2 , x 3 , y ) : ≡ Sub ( x 1 , x 2 , x 3 , y ) ∧ ( ∀ z < y )[ ¬ Sub ( x 1 , x 2 , x 3 , z )] . This lets us use Rosser’s Lemma to eliminate bounded quantifiers in N .