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Proof step 13 from proof div-minus-onediff (previous step)

j0 < j, j0 < 0, i0 * j + j0 < j, i0 * j + j0 < 0, i0 < j + j0, i0 < j, i0 ≤ 0, (i0 * j + j0) + 1 < j, 0 < j + j0, j0 + 1 < j, 1 < j, 0 < j ⊦ j0 = 0, i0 = ((i0 * j + j0) + 1 - j) / j + 1, i0 * j = j * (i0 - 1) + j

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0, 1, i