"Any proof which uses the axiom of choice can be transformed into one that doesn't"?! Lol. If C is the axiom of choice, then C (vacuously) proves C. By your assertion, that proof can be 'transformed' into a proof not using C, which means you can prove C from ZF, which is a contradiction. Really, lol. [[User:Tomkup32|Tomkup32]] 09:26, 9 December 2009 (EST) | "Any proof which uses the axiom of choice can be transformed into one that doesn't"?! Lol. If C is the axiom of choice, then C (vacuously) proves C. By your assertion, that proof can be 'transformed' into a proof not using C, which means you can prove C from ZF, which is a contradiction. Really, lol. [[User:Tomkup32|Tomkup32]] 09:26, 9 December 2009 (EST) |