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Proof step 8 from proof sub-01 (previous step)

i0 < 0, i0 + j ≤ 0, j ≤ 0 ⊦ i0 = 0, i→n(i0) = i→n(i0 + j) - i→n(j)

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: add seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) + i→n(j) = i→n(i + j) subst: [i, j] → [~ i0, ~ j]

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);% ⊦ ~ i + ~ j = ~(i + j); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m + n - m = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ i + 0 = i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: ~ i0, ~ j