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Proof step 10 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 11:
i + i = j0 ∨ i + i + 1 = j0 ⊦ ∃ i0. (i0 + i0) + 1 = j0 ∨ (i0 + i0 + 1) + 1 = j0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;% ⊦ i = j -1 ↔ i +1 = j; (s)