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Proof step 4 from proof ex-oddeven (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i + i = j0 ∨ i + i + 1 = j0 ⊦ ∃ i0. i0 + i0 = j0 + 1 ∨ i0 + i0 = j0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%add-in: ⊦ i + j = i + i0 ↔ j = i0;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;