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Proof step 19 from proof less-mult-div-01 (previous step)

i3 * i1 + i2 * (i + i3) + j0 = 0, 1 < i1 + i2, 0 < i2, 0 < i1, 0 ≤ j0, j0 < i1 + i2, 0 < j1, 0 < i, 0 < i + i3, 0 ≤ i3 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-basic2 inst: name: zero-06 seq: ⊦ ¬ i < 0 ∧ ¬ j < 0 → (0 = i + j ↔ i = 0 ∧ j = 0) subst: [j, i] → [i2 * (i + i3) + j0, i3 * i1] flags: rewrite

Additional information

Used simplifier rules: zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%ls-dist-02: ⊦ i * (j + i0) + j2 < i * (j0 + i0) + j1 ↔ i * j + j2 < i * j0 + j1; from specification int-mult%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i2 * (i + i3) + j0, i3 * i1