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Proof step 10 from proof odd-rec (
previous step
)
2 * j = i0 + 1, 0 ≤ i0 ⊦ ∃ j0. i0 = 2 * j0 + 1
User applied rule exists right
And obtained the following new goals
Proof step 11
:
2 * j = i0 + 1, 0 ≤ i0 ⊦ i0 = 2 * (j - 1) + 1
Rule argument
pos: right 1 terms: j - 1 abort: F discard: T computed: F
Additional information
Used terms: j - 1