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Proof step 9 from proof ex-oddeven (
previous step
)
⊦ ∃ i0. i0 + i0 = (i + i + 1) + 1 ∨ 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; Used terms: i + 1