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Proof step 14 from proof ex-oddeven (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ ∃ i0. i0 + i0 = i + i ∨ 2 + i0 + i0 = i + i + 1

Rule argument

Additional information

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