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Proof step 25 from proof div-eq-zero-01 (previous step)

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

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-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%ls-dist-04: ⊦ i < i * (j + 1) + j1 ↔ 0 < i * j + j1; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%numls-01: ⊦ i + (* j) < i0 + (* j0) ↔ i < i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%zero-gr-add-01: ⊦ 0 ≤ i ∧ j < 0 ∧ j0 ≤ 0 → ¬ 0 < i * j + j0; from specification int-mult% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0, 1, i