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Proof step 11 from proof ex-oddeven (previous step)

i + i = j0 ∨ i + i + 1 = j0 ⊦ ∃ i0. (i0 + i0) + 1 = j0 ∨ 2 + i0 + i0 = j0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information