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1 + 22 = 5one can prove that
1 ( + 22 = 5)For it follows from our premise (by -Abstraction) that:
[z z + 22 = 5]1Independently, by the logic of identity, we know:
([z z + 22 = 5]) = ([z z + 22 = 5])So we may conjoin this fact and the result of -Abstraction to produce:
([z z + 22 = 5]) = ([z z + 22 = 5]) & [z z + 22 = 5]1Then, by existential generalization on the concept [z z + 22 = 5], it follows that:
G[([z z + 22 = 5]) = G & G1]By the definition of membership, we obtain:
1 ([z z + 22 = 5])And, finally, by our Rewrite Rule, we establish what we set out to prove:
1 ( + 22 = 5)
Return to Frege's Logic, Theorem, and Foundations for Arithmetic
First published: June 10, 1998
Content last modified: May 9, 2002