prove: ∃ a b, irrational a ∧ irrational b ∧ ¬ irrational (a^b)

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Formal to Formal Mathematics Benchmark

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Closed cuppajoeman closed 2 years ago

cuppajoeman commented 2 years ago

Thanks to spolu for the bump, that let me complete this proof!

spolu commented 2 years ago

💥