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Proof step 22 from proof div-minus-onediff-01 (previous step)

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

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] → [j0, 1, i] flags:

Additional information

Used simplifier rules: eq-succ: ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%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: j0, 1, i