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Proof step 24 from proof div-div-pred (previous step)

2 * (i1 * 2 + j1) = (i0 * 4 + j2) + 1, abs(j2) < abs(2 * j1 - 1), 4 ≤ abs(2 * j1 - 1), abs(j2) < 4, abs(j1) < 2, i0 * 4 + j2 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, j2 ≤ 0, i0 ≤ 0, i0 * 4 + j2 < 0 ⊦ i1 = i0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
2 * j1 = i2 + 1, 2 * (i1 * 2 + j1) = (i0 * 4 + j2) + 1, abs(j2) < abs(i2), 4 ≤ abs(i2), abs(j2) < 4, abs(j1) < 2, i0 * 4 + j2 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, j2 ≤ 0, i0 ≤ 0, i0 * 4 + j2 < 0 ⊦ i1 = i0

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [2 * j1, 1, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: 2 * j1, 1, i