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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
i = i0 + j, i ≤ j, i ≤ 0, j ≤ 0 ⊦ i→n(i0) = i→n(i) - i→n(j)

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, j, i0] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, j, i0