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Proof step 7 from proof sub-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i0 < 0, i0 + j ≤ 0, j ≤ 0 ⊦ i0 = 0, i→n(i0) = i→n(i0 + j) - i→n(j)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ m - m = 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)