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Proof step 1 from proof sub-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i ≤ j, i ≤ 0, j ≤ 0 ⊦ i→n(i - j) = i→n(i) - i→n(j)

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs