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Proof step 5 from proof ex-two (previous step)

2 * m0 = m ∨ 2 * m0 + 1 = m ⊦ ∃ n. m = 2 * n, ∃ n. m = 2 * n + 1

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information