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Supplement to Frege's Theorem and Foundations for Arithmetic

A More Complex Example

If given the premise that

1+22=5

one can prove that

1ϵ[λzz+22=5]

For it follows from our premise (by λ-Abstraction) that:

[λzz+22=5]1

Independently, by the logic of identity, we know:

ϵ[λzz+22=5]=ϵ[λzz+22=5]

So we may conjoin this fact and the result of λ-Abstraction to produce:

ϵ[λzz+22=5]=ϵ[λzz+22=5]&[λzz+22=5]1

Then, by existential generalization on the concept [λzz+22=5], it follows that:

G[ϵ[λzz+22=5]=ϵG&G1]

And, finally, by the definition of membership, we obtain:

1ϵ[λzz+22=5],

which is what we were trying to prove.

Return to Frege's Theorem and Foundations for Arithmetic

Copyright © 2017 by
Edward N. Zalta <zalta@stanford.edu>

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