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Proof step 12 from proof square-ex (previous step)

(j + 1) * (j + 1) = i0 + 1, 0 < i0, 0 ≤ i0, j * j ≤ i0, 0 < j, 2 * (2 + j) + j * (2 + j) ≤ i0 + 1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: dist-lsplit seq: ⊦ (i + i0) * j = i * j + i0 * j subst: [j, i0, i] → [j + 1, 1, j] flags: rewrite, current

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%le-dist-02: ⊦ i * (j + i0) + j2 ≤ i * (j0 + i0) + j1 ↔ i * j + j2 ≤ i * j0 + j1;%numle-03: ⊦ (* i2) * i + j ≤ i + j0 ↔ (i2 - 1) * i + j ≤ j0;%le-one: ⊦ 1 < j → (i * j ≤ 1 ↔ i ≤ 0);%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%lep-02: ⊦ i + j ≤ i + j0 ↔ j ≤ j0; from specification int-abs%numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs%zero-gels-02: ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i; from specification int-abs Used terms: j + 1, 1, j