ΘρϵηΠατπ

first order language
A first order language ℒ is an infinite collection of distinct symbols, no one of which is contained in another, separated into the following categories
  • Parenthesis: (,)
  • Connectives: ∨,¬
  • Quantifier: ∀
  • Variables, one for each i∈ℕ1: vi
  • Equality symbol: =
  • Constant symbols: A set of symbols
  • Function symbols, for each n∈ℕ1: A set of n-ary function symbols
  • Relation symbols, for each n∈ℕ1: A set of n-ary relation symbols

Since the only thing differing from language to language are its constants, variables, functions and relations then we can denote the language ℒ by (Cℒ,Fℒ,Rℒ), we can denote the set of variables by Vℒ

term
If ℒ is a language, a term of ℒ is a nonempty finite string t of symbols from ℒ such that
  • t is a variable
  • t is a constant
  • t≡ft1t2…tn is a constant where f is an n-ary function symbol if ℒ and each ti is a term of ℒ
formula
If ℒ is a first order language, then a formula of ℒ is a non-empty finite string ϕ of symbols from ℒ such that:
  1. ϕ:≡=t1t2 where t1,t2 are terms of ℒ
  2. ϕ:≡Rt1t2…tn where R is an n-ary relation symbol of ℒ and t1,t2 are terms of ℒ
  3. ϕ:≡(¬α) where α is a formula of ℒ
  4. ϕ:≡(α∨β) where α,β are formulas of ℒ
  5. ϕ:≡(∀v)(α) where v is a variable and α is a formula of ℒ
Atomic Formula
The atomic formulas of ℒ are formulas that satisfy clause 1 or 2 of their definition
language of number theory
We define the language ℒNT=({𝟶},{𝚂,+,⋅,𝙴},{<}) as the language of number theory.
ℒ-Structure
Fix a language ℒ. An ℒ-structure 𝔄 is a non-empty set A, called the universe of 𝔄, such that the following holds:
  • For each constant symbol c of ℒ, we have an element c𝔄 of A
  • For each n-ary function symbol f of ℒ, we have a function f𝔄:An→A
  • For each n-ary relation symbol R of ℒ, we have an n-ary relation R𝔄 on A
Similar to a language we may denote it by (A,{c𝔄:c∈Cℒ},{f𝔄:f∈Fℒ},{R𝔄:R∈Rℒ})
standard number theory structure
Given ℒNT, we define the structure 𝔑:=(ℕ0,{0},{S,+,⋅,E},{<}), where the constant symbol 𝟶 maps to 0 (the element of ℕ0), S is the successor function S(2)=3, + is usual addition 3+3=6, ⋅ multiplication 2⋅4=8, E for exponentiation E(3,2)=9
variable assignment function
If 𝔄 is an ℒ-structure, a variable assignment function is any function of the form s:Vℒ→A
x-modification of an assignment function
If s is a variable assigment function into 𝔄 and x∈Vℒ and a∈A, then s[x|a] is the variable assigment function defined as follows
s[x|a](v):={s(v)if v is a variable other than xaif v is the variable x
and we say that s[x|a] is an x-modification of the assigment function s
An x-modification of s is just like s, except we bind the variable x to the element a of 𝔄's universe
term assignment function
Suppose that 𝔄 is an L-structure and s is a variable assigment function into 𝔄. The function s, called the term assigment function generated by s, is the function with domain consisting of the set of ℒ terms and codomain A defined as follows
  • if t is a variable then s(t)=s(t)
  • if t is the constant c then s(t)=c𝔄
  • if t:≡ft1t2…tn then s(t)=f𝔄(s(t1),s(t2),…,s(tn))
interpretation of a term
Given an ℒ-structure 𝔄, and a term t from ℒ, we say that it's interpretation in 𝔄 is s(t)
numbers in the number theory structure
What is the interpretation in 𝔑 of the term t defined as 𝙴𝚂𝚂𝚂𝟶𝚂𝚂𝟶

The interpretation is s(𝙴𝚂𝚂𝚂𝟶𝚂𝚂𝟶) which is equal to 𝙴𝔄(s(𝚂𝚂𝚂𝟶),s(𝚜𝚜𝟶)), we know that 𝙴𝔄=E (the exponentiation function).

Now considering s(𝚂𝚂𝟶) it becomes S(s(𝚂𝟶)) which becomes S(S(s(𝟶))) and since s(𝟶)=𝟶𝔄=0 (the natural number 0), then s(𝚂𝚂𝟶)=0+1+1=2, similarly we can find that s(𝚂𝚂𝚂𝟶)=3

From the above paragrpah we know that E(s(𝚂𝚂𝚂𝟶),s(𝚜𝚜𝟶)) becomes E(3,2) which equals 9, so therefore the interpretation of 𝙴𝚂𝚂𝚂𝟶𝚂𝚂𝟶 in 𝔑 is 9.

A Free Varaible in a Formula
Suppose that v is a varaible and ϕ is a formula, then v is free in ϕ iff exactly one of the following holds:
  1. ϕ is atomic and v occurs in α
  2. ϕ:≡(¬α) and v is free in α
  3. ϕ:≡(α∨β) and v is free in at least one of α or β
  4. ϕ:≡(∀u)(α) and v is not u and v is free in α
For All Makes a Variable not Free
Free Variable For All Easy
Show that x is not free in ϕ:≡(∀x)(x=x)
Free Variable For All Medium
Show that in the following formula ϕ defined as (∀v2)(¬((∀v3)((v1=S(v2)∨v3=v2)))) v1 is free in ϕ, but v2 and v3 are not

We'll start by showing that v2 is free in phi