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Proof step 2 from proof ex-oddeven (previous step)

⊦ ∃ i. i + i = 0 ∨ i + i + 1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ∃ i. i = 0 ∨ i + i + 1 = 0

Rule argument

Additional information

Used simplifier rules: zero-07: ⊦ i + i = 0 ↔ i = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;