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Proof step 2 from proof sub-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i ≤ j, j ≤ 0, i ≤ 0 ⊦ i→n(i - j) = i→n(i) - i→n(j)

Rule argument

Additional information

Used simplifier rules: forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs