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Proof step 4 from proof sub-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
0 ≤ i, 0 ≤ j ⊦ i→n(i) + i→n(j) = i→n(i + j)
Proof step 6:
¬(0 ≤ i0 ∧ 0 ≤ j), j ≤ 0, i0 + j ≤ 0, i0 ≤ 0 ⊦ i→n(i0) = i→n(i0 + j) - i→n(j)

Rule argument

spec: intnat inst: name: add seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) + i→n(j) = i→n(i + j) subst: [j, i] → [j, i0] flags: rewrite, current

Additional information

Used simplifier rules: eq-01: ⊦ m + n - m = n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: j, i0