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Proof step 8 from proof ls-fin-add (previous step)

n + i→n(i) = m0 + 1, m0 = n + i→n(i) -1, 0 < n→i(n) + i, n→i(m0) + 1 < n→i(n) + i, n→i(m0) < n→i(n) + i, 0 ≤ i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < n→i(n) + i, n→i(i→n(i) - 1) < i, n→i(i→n(i) - 1) + 1 < i, 0 < i, 0 ≤ i ⊦

Rule argument

Additional information

Used simplifier rules: num-i2n-le-02: 0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i;%n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%pos-int: ⊦ n→i(m) < j → 0 < j;%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j < i + i0 ↔ j < i0; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n -1 = n - 1; (s)