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nomadreid

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- #1

nomadreid

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I don't know if any of the Robinson axioms can be removed while still leaving the incompleteness theorem in place. My guess would be No, but I've never seen any investigation of it.

- #3

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1. ##\Delta_p + \Delta_q = \Delta_{p+q}##;

2. ##\Delta_p \times \Delta_q = \Delta_{p \times q}##;

3. For each ##p, q## such that ##p \not =q##, ##\Delta_p \not = \Delta_q##;

4. ##x \leq \Delta_p \rightarrow x = \Delta_0 \vee \dots \vee x = \Delta_p##;

5. ##x \leq \Delta_p \vee \Delta_p \leq x##.

On pages 53-54, they show that all these axioms are derivable from

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nomadreid

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You're welcome. Fortunately, the book is available as a very cheap Dover paperback, so it won't cost you much!

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nomadreid

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