∆-Definability of Sequence-Coding Operations Let P1 = 2, P2 = 3, ..., Pi

∆-Definability of Sequence-Coding Operations Let P1 = 2, P2 = 3, ..., Pi

∆-Definability of Sequence-Coding Operations th Let p1 = 2, p2 = 3,..., pi = i prime number. N<N is the set of finite sequences of natural numbers. Sequence-coding function hi : N<N ! N is defined by k Y ai+1 ha1; : : : ; aki := pi : i=1 If a = ha1; : : : ; aki, then jaj = k and( a)i = ai. All of the sequence-coding operations are ∆-definable (hence, representable). We have∆-formulas: Divides(y; x): ≡ (9z ≤ x)[x = y · z]; Prime(x): ≡ S0 < x ^ (8y ≤ x)[Divides(y; x) ! (y = 1 _ y = x)] We have∆-formulas: Divides(y; x): ≡ (9z ≤ x)[x = y · z]; Prime(x): ≡ S0 < x ^ (8y ≤ x)[Divides(y; x) ! (y = 1 _ y = x)] | {z } :Divides(y; x) _ (y = 1 _ y = x) Remark: Technically, :Divides(y; x) is not a∆-formula. However, it is equivalent to a∆-formula. The set 2 PrimePair := f(pi; pi+1): i ≥ 1g ⊆ N is defined by the∆-formula PrimePair(x; y):≡ Prime(x) ^ Prime(y) ^ x < y ^ (8z < y)[Prime(z) ! z ≤ x]: The set 2 PrimePair := f(pi; pi+1): i ≥ 1g ⊆ N is defined by the∆-formula PrimePair(x; y):≡ Prime(x) ^ Prime(y) ^ x < y ^ (8z < y)[Prime(z) ! z ≤ x]: The set <N Codenumber = fh a1; : : : ; aki :(a1; : : : ; ak) 2 N g ⊆ N | {z } a1+1 a2+1 ak+1 p1 p2 ···pk is defined by the∆-formula Codenumber(c):≡ Divides(SS0; c) ^ (8z < c)(8y < z) PrimePair(y; z) ^ Divides(z; c) ! Divides(y; c): This formula expresses: \c is divisible by2 and for every prime pair( pi; pi+1), if pi+1 divides c, so pi divides c." Continuing, we define the set Yardstick := fh0; 1; 2; : : : ; k − 1i : k 2 Ng | {z } 1 2 3 k 2 3 5 :::(pk) by the∆-formula Yardstick(x):≡ Divides(2; x) ^ :Divides(4; x) ^ (8y ≤ x)(8z ≤ x)(8i < x) PrimePair(y; z) ^ Divides(z; x) : ! Divides(y E i; x) $ Divides(z E Si; x) |{z} | {zi+1 } yi z We next define the set IthPrime := f(i; pi): i 2 N≥1g by the∆-formula 2Yardstick(x) ^ 3 6Divides(yi; x) ^ 7 IthPrime(i; y):≡ Prime(y) ^ (9x ≤ some term) 4 5 : :Divides(yi+1; x) Question: What term suffices for this bounded quantifier? We next define the set IthPrime := f(i; pi): i 2 N≥1g by the∆-formula 2Yardstick(x) ^ 3 2 i 6Divides(yi; x) ^ 7 IthPrime(i; y):≡ Prime(y) ^ (9x ≤ y ) 4 5 : :Divides(yi+1; x) Question: What term suffices for this bounded quantifier? 1 2 i Answer: We want x to be the i-th yardstick number( p1) (p2) ::: (pi) . This is at most 2 3 i (i+1) i2 (pi)(pi) (pi) ··· (pi) = (pi) 2 ≤ (pi) : i2 Therefore, if y = pi, it suffices to take x ≤ y . We next define the set IthPrime := f(i; pi): i 2 N≥1g by the∆-formula 2Yardstick(x) ^ 3 2 i 6Divides(yi; x) ^ 7 IthPrime(i; y):≡ Prime(y) ^ (9x ≤ y ) 4 5 : :Divides(yi+1; x) Question: What term suffices for this bounded quantifier? 1 2 i Answer: We want x to be the i-th yardstick number( p1) (p2) ::: (pi) . This is at most 2 3 i (i+1) i2 (pi)(pi) (pi) ··· (pi) = (pi) 2 ≤ (pi) : i2 Therefore, if y = pi, it suffices to take x ≤ y . i i2 Alternatively, since pi ≤ (i + 1) (easy fact), we could instead use x ≤ (i + 1) . We next define the set IthPrime := f(i; pi): i 2 N≥1g by the∆-formula 2Yardstick(x) ^ 3 i·i 6Divides(yi; x) ^ 7 IthPrime(i; y):≡ Prime(y) ^ (9x ≤ y ) 4 5 : :Divides(yi+1; x) Exercise. Convince yourself that IthPrime(i; y) indeed defines IthPrime. That is, show that • N j= IthPrime(k; pk) for every k ≥ 1, • N j= :IthPrime(a; b) whenever a = 0 or b =6 pa. We next define the set IthPrime := f(i; pi): i 2 N≥1g by the∆-formula 2Yardstick(x) ^ 3 i·i 6Divides(yi; x) ^ 7 IthPrime(i; y):≡ Prime(y) ^ (9x ≤ y ) 4 5 : :Divides(yi+1; x) Remark. The set IthPrime ⊆ N2 corresponds to the function N ! N defined by i 7! pi. Therefore, we say that the formula IthPrime(i; y) defines the function i 7! pi. Continuing, the set k Length := (ha1; : : : ; aki; k): k ≥ 1 and (a1; : : : ; ak) 2 N is defined by the∆-formula Length(c; `):≡ Codenumber(c) IthPrime(`; y) ^ Divides(y; c) ^ (9y ≤ c) : ^ (8z ≤ c)[PrimePair(y; z) !:Divides(z; c)] Continuing, the set k Length := (ha1; : : : ; aki; k): k ≥ 1 and (a1; : : : ; ak) 2 N is defined by the∆-formula Length(c; `):≡ Codenumber(c) IthPrime(`; y) ^ Divides(y; c) ^ (9y ≤ c) : ^ (8z ≤ c)[PrimePair(y; z) !:Divides(z; c)] The set k IthElement := (aj; j; ha1; : : : ; aki) : 1 ≤ j ≤ k and (a1; : : : ; ak) 2 N is defined by the∆-formula IthElement(e; i; c):≡ Codenumber(c) ^ (9y ≤ c) IthPrime(i; y) ^ Divides(ySe; c) ^ :Divides(ySSe; c): ∆-Definability of Sequence-Coding Operations For practice, try writing a∆-formula that defines the set Concatenation ⊆ 3 N of triples of the form( ha1; : : : ; aki; hb1; : : : ; b`i; ha1; : : : ; ak; b1; : : : ; b`i). Godel¨ Numbers of Terms and Formulas We assign a unique number to each symbol in LNT as follows: : 1 · 15 _ 3 E 17 8 5 < 19 =7 ( 21 09 ) 23 S 11 vi 2i + 13 Suppose s :≡ s1 : : : sn is a string of symbols, which constituting a well-formed term or formula of LNT . Naively, we could encode s by the number h#(s1);:::; #(sn)i where#( si) is the number corresponding to the symbol si. However, it much better to encode s according to the inductive type of terms and formulas. Def 5.7.1. For each term t and formula ', the G¨odelnumbers ptq and p'q are defined as follows: p:αq = h1; pαqi p+t1t2q = h13; pt1q; pt2qi p(α _ β)q = h3; pαq; pβqi p· t1t2q = h15; pt1q; pt2qi p(8vi)(α)q = h5; pviq; pαqi pEt1t2q = h17; pt1q; pt2qi p=t1t2q = h7; pt1q; pt2qi p<t1t2q = h19; pt1q; pt2qi p0q = h9i pviq = h2ii: pStq = h11; ptqi Def 5.7.1. For each term t and formula ', the G¨odelnumbers ptq and p'q are defined as follows: p:αq = h1; pαqi p+t1t2q = h13; pt1q; pt2qi p(α _ β)q = h3; pαq; pβqi p· t1t2q = h15; pt1q; pt2qi p(8vi)(α)q = h5; pviq; pαqi pEt1t2q = h17; pt1q; pt2qi p=t1t2q = h7; pt1q; pt2qi p<t1t2q = h19; pt1q; pt2qi p0q = h9i pviq = h2ii: pStq = h11; ptqi Obs. ptq and p'q are never divisible by7. (Why?) Def 5.7.1. For each term t and formula ', the G¨odelnumbers ptq and p'q are defined as follows: p:αq = h1; pαqi p+t1t2q = h13; pt1q; pt2qi p(α _ β)q = h3; pαq; pβqi p· t1t2q = h15; pt1q; pt2qi p(8vi)(α)q = h5; pviq; pαqi pEt1t2q = h17; pt1q; pt2qi p=t1t2q = h7; pt1q; pt2qi p<t1t2q = h19; pt1q; pt2qi p0q = h9i pviq = h2ii: pStq = h11; ptqi Example. p=0S0q = h7; p0q; pS0qi = h7; h9i; h11; h9iii 12 1025 = h7; 210; h11; 210ii = 28310255(2 3 +1): Notice how fast SSSS0 grows: p q 10 12 21232 12 21232 3 pSSSS0q = h11; h11; h11; h11; h9iiiii =2 3 Next Steps (Section 5.8) ∆-definability of sets Terms := fptq : terms tg = fa 2 N : a = ptq for some term tg; Formulas := fp'q : formulas 'g = fa 2 N : a = p'q for some formula 'g:.

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