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Proof step 15 from proof ls-fin-add (previous step)

i→n(i0) = 2 + m, 2 + n→i(m) < i0, 0 < n→i(n) + i0, 1 < n→i(n) + i0, 0 < i0, 1 < i0 ⊦ i0 = 0, i0 = 1, ∃ n0. 0 < n→i(n0) + i0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: 0 abort: F discard: T computed: F

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: 0