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Proof step 3 from proof ls-fin-add (previous step)

n→i(m) < n→i(n) + i, 0 < n→i(n) + i, n + i→n(i) ≤ m, 0 ≤ i ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
0 < n→i(n) + i, n→i(m0 +1) < n→i(n) + i, 0 ≤ i, n + i→n(i) ≤ m0 +1, ∀ n, i. ¬(n→i(m0) < n→i(n) + i ∧ 0 < n→i(n) + i ∧ n + i→n(i) ≤ m0 ∧ 0 ≤ i) ⊦

Rule argument

indvar: m

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%nat freely generated by 0, +1;;% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2