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

0 ≤ i, n→i(m0) < n→i(n) + i, n + i→n(i) ≤ m0, 0 < n→i(n) + i, (∃ n, i. n→i(m0) ≤ n→i(n) + i ∧ n + i→n(i) < m0 ∧ 0 ≤ i) → m0 = 0 ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 6:
0 ≤ i, n→i(m0) < n→i(n) + i, n + i→n(i) ≤ m0, 0 < n→i(n) + i ⊦ ∃ n, i. n→i(m0) ≤ n→i(n) + i ∧ n + i→n(i) < m0 ∧ 0 ≤ i

Rule argument

pos: left 5

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2