Exercises #2 – Independence of Axioms

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

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

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

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

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

[latex]A_i[/latex]

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

[latex]\blacksquare[/latex]

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

[latex]\blacksquare[/latex]

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