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Proof step 1 from proof sub (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j ≤ i, 0 ≤ i, 0 ≤ j ⊦ 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%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs