ΘρϵηΠατπ

first order language
A first order language 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 i1: vi
  • Equality symbol: =
  • Constant symbols: A set of symbols
  • Function symbols, for each n1: A set of n-ary function symbols
  • Relation symbols, for each 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 by (C,F,R), we can denote the set of variables by V

term
If is a language, a term of is a nonempty finite string t of symbols from such that
  • t is a variable
  • t is a constant
  • tft1t2tn is a constant where f is an n-ary function symbol if and each ti is a term of
formula
If is a first order language, then a formula of is a non-empty finite string ϕ of symbols from such that:
  1. ϕ:=t1t2 where t1,t2 are terms of
  2. ϕ:Rt1t2tn where R is an n-ary relation symbol of and t1,t2 are terms of
  3. ϕ:(¬α) where α is a formula of
  4. ϕ:(αβ) where α,β are formulas of
  5. ϕ:(v)(α) where v is a variable and α is a formula of
Atomic Formula
The atomic formulas of are formulas that satisfy clause 1 or 2 of their definition
language of number theory
We define the language NT=({𝟶},{𝚂,+,,𝙴},{<}) as the language of number theory.
-Structure
Fix a language . An -structure 𝔄 is a non-empty set A, called the universe of 𝔄, such that the following holds:
  • For each constant symbol c of , we have an element c𝔄 of A
  • For each n-ary function symbol f of , we have a function f𝔄:AnA
  • For each n-ary relation symbol R of , we have an n-ary relation R𝔄 on A
Similar to a language we may denote it by (A,{c𝔄:cC},{f𝔄:fF},{R𝔄:RR})
standard number theory structure
Given NT, we define the structure 𝔑:=(0,{0},{S,+,,E},{<}), where the constant symbol 𝟶 maps to 0 (the element of 0), S is the successor function S(2)=3, + is usual addition 3+3=6, multiplication 24=8, E for exponentiation E(3,2)=9
variable assignment function
If 𝔄 is an -structure, a variable assignment function is any function of the form s:VA
x-modification of an assignment function
If s is a variable assigment function into 𝔄 and xV and aA, 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 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:ft1t2tn then s(t)=f𝔄(s(t1),s(t2),,s(tn))
interpretation of a term
Given an -structure 𝔄, and a term t from , 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:(αβ) 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