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Proof step 9 from proof ex-oddeven (previous step)

⊦ ∃ i0. i0 + i0 = 2 + i + i ∨ i0 + i0 = i + i + 1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i + 1 abort: F discard: F computed: F

Additional information

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