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Ordinal Subtraction Defined by Recursion

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Please see the attached problem. I have attached some notes in case these are needed.

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Proposition 2.7.9: For any pair of ordinals alpha and beta such that alpha <= beta, there is a unique ordinal gamma such that alpha + gamma = beta, so we can write (unambiguously) gamma = beta - alpha when beta >= alpha.

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The given ordinal subtraction operation (defined by recursion):

alpha - alpha = 0

(beta^+) - alpha = (beta - alpha)^+

gamma - alpha = sup{beta - alpha: beta < gamma} when gamma is a limit ordinal.

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Solution Preview

Please see the attached file for a solution to the problem.

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Update: As you will see in the .pdf file that contains your Notes, the process of taking proofs that break down into the "0 case", the "successor case", and the "limit ordinal case" is known as ...

Solution Summary

A complete, detailed proof that the given ordinal subtraction operation (defined by recursion) coincides with the definition of ordinal subtraction given in Proposition 2.7.9 is provided.

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