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Proof step 5 from proof ls-fin-add (previous step)

0 < n→i(n) + i, 0 ≤ i, n→i(m0) + 1 < n→i(n) + i, n + i→n(i) ≤ m0 + 1, ∀ n0, i0. ¬ n→i(m0) < n→i(n0) + i0 ∨ ¬ n0 + i→n(i0) ≤ m0 ∨ ¬ 0 < n→i(n0) + i0 ∨ ¬ 0 ≤ i0 ⊦

User applied rule all left

And obtained the following new goals

Proof step 6:
0 < n→i(n) + i, 0 ≤ i, n→i(m0) + 1 < n→i(n) + i, n + i→n(i) ≤ m0 + 1, ¬ n→i(m0) < n→i(n) + i ∨ ¬ n + i→n(i) ≤ m0 ∨ ¬ 0 < n→i(n) + i ∨ ¬ 0 ≤ i ⊦

Rule argument

pos: left 5 terms: n, i abort: F discard: T computed: T

Additional information

Used terms: n, i