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Proof step 4 from proof ls-fin-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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 ⊦

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2