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Proof step 1 from proof sub-same (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs