Supplement to Frege's Theorem and Foundations for Arithmetic

A More Complex Example

If given the premise that


one can prove that

1 ∈ ε[λz z+22 =5]

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

z z+22=5]1

Independently, by the logic of identity, we know:

ε[λz z+22=5]   =   ε[λz z+22=5]

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

ε[λz z+22=5] = ε[λz z+22=5]   &   [λz z+22=5]1

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

G[ε[λz z+22=5] = εG   &   G1]

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

1 ∈ ε[λz z+22=5],

which is what we were trying to prove.

Return to Frege's Theorem and Foundations for Arithmetic

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Edward N. Zalta <>

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