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Proof step 8 from proof ex-oddeven (previous step)

i + i + 1 = j0 ⊦ ∃ i0. i0 + i0 = j0 + 1 ∨ i0 + i0 = j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ ∃ i0. i0 + i0 = 2 + i + i ∨ i0 + i0 = i + i + 1

Rule argument

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;