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Supplement to Frege's Logic, Theorem, and Foundations for Arithmetic

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1+2one can prove that^{2}=5

1 ∈ ε[λFor it follows from our premise (by λ-Abstraction) that:zz+2^{2}=5]

[λIndependently, by the logic of identity, we know:zz+2^{2}=5]1

ε[λSo we may conjoin this fact and the result of λ-Abstraction to produce:zz+2^{2}=5] = ε[λzz+2^{2}=5]

ε[λThen, by existential generalization on the concept [λzz+2^{2}=5] = ε[λzz+2^{2}=5] & [λzz+2^{2}=5]1

∃And, finally, By the definition of membership, we obtain:G[ε[λzz+2^{2}=5] = εG&G1]

1 ∈ ε[λwhich is what we were trying to prove.zz+2^{2}=5],

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