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

i < j ⊦ i + 1 = j, abs(j - (i + 1) + 1) < abs(j - i)

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:
j = i0 + (i + 1) + 1, i < j ⊦ i + 1 = j, abs(i0) < abs(j - i)

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, (i + 1) + 1, i0] flags: rewrite, current

Additional information

Used terms: j, (i + 1) + 1, i0