# Exercises #2 – Independence of Axioms

These were pulled from Exercises for $\S$2 of Chapter I in Bourbaki’s set theory book.

#### Exercise 1.

Let $\mathfrak{T}$ be a theory, let $A_1,A_2,…,A_n$ be its explicit axioms and $a_1,a_2,…,a_h$ its constants.

1. Let $\mathfrak{T}'$ be the theory with signs and schemes are the same as those of $\mathfrak{T}$, and whose explicit axioms are $A_1,A_2,…,A_{n-1}$. The axiom $A_n$ is said to be independent of the other axioms of $\mathfrak{T}$ if $\mathfrak{T}'$ is not equivalent to $\mathfrak{T}$. Show that $A_n$ is independent if and only if $A_n$ is not a theorem of $\mathfrak{T}'$.
2. Let $\mathfrak{T}^{\prime\prime}$ be the theory with signs and schemes are the same as those of $\mathfrak{T}$. Let $T_1,T_2,…,T_h$ be terms of $\mathfrak{T}$ such that
$(T_1|a_1)(T_2|a_2)…(T_h|a_h)A_i$

is a theorem of $\mathfrak{T}^{\prime\prime}$ for $i=1,2,…,n-1$ and such that $\neg (T_1|a_1)(T_2|a_2)…(T_h|a_h)A_n$ is a theorem of $\mathfrak{T}^{\prime\prime}$. Then show either $A_n$ is independent of the other axioms of $\mathfrak{T}$, or else $\mathfrak{T}^{\prime\prime}$ is contradictory.

#### Solution for a.

We will show the contrapositive. Note that every explicit axiom in $\mathfrak{T}'$ is a theorem in $\mathfrak{T}'$ since the formative construction

$A_i$

(where $i=1,2,…,n-1$) contains $A_i$ and satisfies the conditions of a proof in $\mathfrak{T}'$. Hence, every explicit axiom except $A_n$ of $\mathfrak{T}$ is a theorem in $\mathfrak{T}'$. Since the signs and schemes are the same in both theories, $\mathfrak{T}'$ is stronger than $\mathfrak{T}$ if and only if $A_n$ is a theorem in $\mathfrak{T}'$. Since every explicit axiom of $\mathfrak{T}'$ is an explicit axiom (hence a theorem) in $\mathfrak{T}$, $\mathfrak{T}$ is stronger than $\mathfrak{T}'$. We thus conclude $\mathfrak{T}'$ is equivalent to $\mathfrak{T}$ if and only if $A_n$ is a theorem in $\mathfrak{T}'$, i.e. $A_n$ is not independent if and only if $A_n$ is a theorem in $\mathfrak{T}'$.

$\blacksquare$

#### Solution for b.

By the previous part, $A_n$ is not independent if and only if $A_n$ is a theorem in $\mathfrak{T}^{\prime\prime}$. Hence, if $A_n$ is not independent, then $(T_1|a_1)(T_2|a_2)…(T_h|a_h)A_n$ is a theorem by I.2.4.C5 in the book. But the negation is also a theorem by hypothesis in $\mathfrak{T}^{\prime\prime}$ implying $\mathfrak{T}^{\prime\prime}$ is contradictory. Either $A_n$ is independent or not, so we are done.

$\blacksquare$

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