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Proof step 3 from proof abssub-05 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, j, i1] flags: rewrite, current

Additional information

Used terms: i, j, i1