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Proof step 2 from proof abs-ls-02 (previous step)

abs(i - j) < abs(j0 - j) ⊦ j0 < j, i < j0, i < j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ i0 = i - j ↔ i = i0 + j
Proof step 4:
i = i0 + j, abs(i0) < abs(j0 - j) ⊦ j0 < j, i < j0, i < 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: rewrite, current

Additional information

Used terms: i, j, i0