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Proof step 10 from proof modlsp-03 (previous step)

abs(j1) < abs(i1 - j), i1 + j1 < abs(i1 - j), i0 < i, 0 < j, j * i + i1 + j1 - j1 < j * i0, abs(j1) < j, i1 + j1 < j, j * i + i1 < j * i0, abs(j1) < abs(i1), i1 + j1 < abs(i1), j ≤ abs(i1 - j), j ≤ abs(i1), 0 ≤ i1 + j1 ⊦ i1 = 0, i1 = j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
i1 = i2 + j, i1 + j1 < abs(i1), abs(j1) < abs(i1), j * i + i1 < j * i0, i1 + j1 < j, abs(j1) < j, j * i + i1 + j1 - j1 < j * i0, 0 < j, i0 < i, i1 + j1 < abs(i2), abs(j1) < abs(i2), 0 ≤ i1 + j1, j ≤ abs(i1), j ≤ abs(i2) ⊦ i1 = j, i1 = 0

Rule argument

spec: int-abs name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i1, j, i2]

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i1, j, i2