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Proof step 3 from proof sub-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
j ≤ 0, i0 + j ≤ 0, i0 ≤ 0 ⊦ i→n(i0) = i→n(i0 + j) - i→n(j)

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs